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DIFFERENTIAL  AND   INTEGRAL   CALCULUS 


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THE  MACMILLAN  COMPANY 

NEW  YORK    •    BOSTON    •    CHICAGO 
ATLANTA  •    SAN  FRANCISCO 

MACMILLAN  &  CO.,  Limited 

LONDON  •    BOMBAY  •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  Ltd. 

TORONTO 


A  FIRST   COURSE 


IN    THE 


DIFFERENTIAL  AND  INTEGRAL 
CALCULUS • 


BY 


WILLIAM   F.   OSGOOD,   PH.D. 

PROFESSOR   OF   MATHEMATICS   IN   HARVARD    UNIVERSITY 


REVISED  EDITION 


Neu)  gorfc 

THE   MACMILLAN   COMPANY 

1909 

All  rights  reserved 


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Copyright,  1907,  1909, 
By   THE   MACMILLAN   COMPANY. 


Set  up  and  electrotyped.     Published  October,  1907.     Reprinted 
April,  June,  1908. 

New  edition  with  additions,  February,  1909. 


Norijjooto  $regs 

J.  S.  Cushing  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

The  treatment  of  the  calculus  that  here  follows  is  based  on 
the  courses  which  I  have  given  in  this  subject  in  Harvard  Col- 
lege for  a  number  of  years  and  corresponds  in  its  main  outlines 
to  the  course  as  given  by  Professor  B.  0.  Peirce  in  the  early 
eighties.  The  introduction  of  the  integral  as  the  limit  of  a 
sum  at  an  early  stage  is  due  to  Professor  Byerly,  who  made 
this  important  change  more  than  a  dozen  years  ago.  Professor 
Byerly,  moreover,  was  a  pioneer  in  this  country  in  teaching  the 
calculus  by  means  of  problems,  his  work  in  this  direction  dat- 
ing from  the  seventies. 

The  chief  characteristics  of  the  treatment  are  the  close  touch 
between  the  calculus  and  those  problems  of  physics,  including 
geometry,  to  which  it  owed  its  origin ;  and  the  simplicity  and 
directness  with  which  the  principles  of  the  calculus  are  set 
forth.  It  is  important  that  the  formal  side  of  the  calculus 
should  be  thoroughly  taught  in  a  first  course,  and  great  stress 
has  been  laid  on  this  side.  But  nowhere  do  the  ideas  that 
underlie  the  calculus  come  out  more  clearly  than  in  its  appli- 
cations to  curve  tracing  and  the  study  of  curves  and  surfaces, 
in  definite  integrals  with  their  varied  applications  to  physics 
and  geometry,  and  in  mechanics.  For  this  reason  these  sub- 
jects have  been  taken  up  at  an  early  stage  and  illustrated  by 
many  examples  not  usually  found  in  American  text-books* 

It  is  exceedingly  difficult  to  cover  in  a  first  course  in  the  cal- 
culus all  the  subjects  that  claim  a  place  there.  Some  teachers 
will  wish  to  see  a  fuller  treatment  of  the  geometry  of  special 

*  Professor  Campbell's  book :  The  Elements  of  the  Differential  and 
Integral  Calculus,  Macmillan,  1904,  in  its  excellent  treatment  of  the  inte- 
gral as  the  limit  of  a  sum,  is  a  notable  exception. 

v 

38351 


vi  PREFACE 

curves  than  I  have  found  room  for.  But  I  beg  to  call  attention 
to  the  importance  of  the  subject  of  functions  of  several  vari- 
ables and  the  elements  of  the  geometry  of  surfaces  and  twisted 
curves  for  all  students  of  the  calculus.  This  subject  ought  not 
to  be  set  completely  aside,  to  be  taken  up  in  the  second  course 
in  the  calculus,  to  which,  unfortunately,  too  few  of  those  who 
take  the  first  course  proceed.  Only  a  slight  knowledge  of  par- 
tial differentiation  is  here  necessary.  .  It  has  been  my  practice 
to  take  up  in  four  or  five  lectures  near  the  end  of  the  first  year 
as  much  about  the  latter  subject  as  is  contained  in  §§  3-9  of 
Chap.  XIV,  omitting  the  proofs  in  §§  7-9,  but  laying  stress  on 
the  theorems  of  these  paragraphs  and  illustrating  them  by  such 
examples  as  those  given  in  the  text ;  and  to  proceed  then  to 
the  simpler  applications  of  Chap.  XV.  Thus  the  way  is  pre- 
pared for  a  thorough  treatment  of  partial  differentiation,  a  sub- 
ject important  alike  for  the  student  of  pure  and  of  applied 
mathematics.  This  subject  was  given  in  the  older  English 
text-books  in  such  a  manner  that  the  student  who  worked 
through  their  exercises  was  able  to  deal  with  the  problems  that 
arise  in  practice.  But  modern  text-books  in  the  English  lan- 
guage are  inferior  to  their  predecessors  in  this  respect.* 

Multiple  integrals  are  usually  postponed  for  a  second  course, 
and  when  they  are  taken  up,  some  of  the  things  that  it  is  most 
important  to  say  about  them  are  omitted  in  the  text-books.  It 
is  the  conception  of  the  double  and  triple  integral  in  its  relation 
to  the  formulation  of  such  physical  ideas  as  the  moment  of  iner- 
tia and  the  area  of  a  surface  that  needs  to  be  set  in  the  fore- 
front of  the  course  in  the  calculus.  And  the  theorem  that  such 
an  integral  can  be  computed  by  a  succession  of  simple  integra- 
tions (the  iterated  integrals)  should  appear  as  a  tool,  as  a  de- 
vice for  accomplishing  a  material  end.  The  conception,  then, 
of  the  double  integral,  its  application  to  the  formulation  of 
physical  concepts,  and  its  evaluation  are  the  things  with  which 

*  I  have  here  to  except  Goursat-Hedrick,  A  Course  in  Mathematical 
Analysis,  vol.  I ;  Ginn  &  Co.,  Boston,  1904. 


PREFACE  vii 

Chap.  XVIII  deals.  In  Chap.  XIX  the  triple  integral  is  ex- 
plained by  analogy  and  computed,  the  analytical  justification 
being  left  for  those  who  are  going  to  specialize  in  analysis. 

The  solution  of  numerical  equations  by  successive  approxi- 
mations and  other  methods,  illustrated  geometrically,  and  the 
computation  of  areas  by  Simpson's  Rule  and  Amsler's  planim- 
eter  are  taken  up  in  Chap.  XX.  In  an  appendix  the  ordinary 
definition  of  the  logarithm  is  justified  and  it  is  shown  that  this 
function  and  its  inverse,  the  exponential  function,  are  con- 
tinuous. 

The  great  majority  of  problems  in  the  calculus  have  come 
down  to  us  from  former  generations,  the  Tripos  Examinations 
and  the  older  English  text-books  having  contributed  an  im- 
portant share.*  For  the  newer  problems  I  am  indebted  in 
great  measure  to  old  examination  papers  set  by  Professor 
Byerly  and  by  Professor  B.  0.  Peirce,f  and  to  recent  Ameri- 
can text-books.  It  is  not  possible  to  acknowledge  each  time 
the  author,  even  in  the  case  of  the  more  recent  problems,  but 
I  wish  to  cite  at  least  a  few  of  the  sources  in  detail.  I  am  in- 
debted to  Campbell  $  for  Ex.  4,  p.  181;  to  Granville  §  for  Ex. 
45,  p.  108 ;  to  Greenhill  ||  for  Ex.  16,  p.  188 ;  and  to  Osborne  IF 
for  Ex.  43  on  p.  107. 

*  In  particular,  Williamson,  An  Elementary  Treatise  on  the  Differen- 
tial Calculus,  University  Press,  Dublin,  and  Todhunter,  A  Treatise  on 
the  Integral  Calculus,  Macmillan. 

t  Many  of  these  problems  have  been  collected  and  published,  with  others, 
by  Professor  Byerly  in  his  Problems  in  Differential  Calculus;  Ginn  &  Co., 
Boston,  1895.  With  the  kind  permission  of  the  author  I  have  drawn 
freely  from  this  source. 

I  Campbell,  I.e.,  Chaps.  XXXVI  and  XXXVII. 

§  Granville,  Elements  of  the  Differential  and  Integral  Calculus,  p.  129, 
Ex.  47;  Ginn  &  Co.,  Boston,  1904. 

II  Greenhill,  A  Treatise  on  Hydrostatics,  p.  318 ;  Macmillan,  1894. 

H  Osborne,  Differential  and  Integral  Calculus,  p.  129,  Ex.  33  ;  D.  C. 
Heath  &  Co.,  Boston,  revised  edition,  1906.  This  book  contains  an  espe- 
cially large  collection  of  exercises. 


viii  PREFACE 

In  choosing  illustrative  examples  to  be  worked  in  the  text  I 
have  taken  so  far  as  possible  the  same  examples  that  my  prede- 
cessors have  used,  in  order  not  to  reduce  further  the  fund  of 
good  examples  for  class-room  purposes  by  publishing  solutions 
of  the  same.  It  is  in  the  interest  of  good  instruction  that 
writers  of  text-books  observe  this  principle. 

As  regards  the  time  required  to  cover  the  course  here  pre- 
sented, I  would  say  that  without  the  aid  of  text-books  which  I 
could  follow  at  all  closely,  I  have  repeatedly  taken  up  what  is 
here  given  in  about  one  hundred  and  thirty-five  lectures,  ex- 
tending over  a  year  and  a  half,  three  lectures  a  week.  The 
time  thus  corresponds  roughly  to  a  five-hour  course  extending 
throughout  one  year. 

To  Mr.  H.  D.  Gaylord,  who  has  given  me  much  assistance 
in  reading  the  proof,  and  to  Dr.  W.  H.  Eoever  and  Mr.  Dunham 
Jackson,  who  have  aided  me  with  the  figures,  I  wish  to  ex- 
press my  appreciation  of  their  kindness. 

Cambridge, 
September  12,  1907. 


CONTENTS 

CHAPTER  I 
Introduction 

ART.  PAGE 

1.  Functions 1 

2.  Slope  of  a  Curve 5 

CHAPTER   II 

Differentiation  of  Algebraic  Functions.     General 
Theorems 

1.  Definition  of  the  Derivative           .        . 9 

2.  Differentiation  of  xn 11 

3.  Derivative  of  a  Constant 13 

4.  General  Formulas  of  Differentiation 13 

6.   Three  Theorems  about  Limits 15 

6.  General  Formulas  of  Differentiation,  Concluded  ....  21 

7.  Differentiation  of  Radicals 25 

8.  Continuation  ;  xn,  n  Fractional 27 

9.  Differentiation  of  Algebraic  Functions 34 


CHAPTER   III 

Applications 

1.  Tangents  and  Normals .37 

2.  Maxima  and  Minima 39 

3.  Continuation  ;  Auxiliary  Variables 43 

4.  Velocity 46 

5.  Increasing  and  Decreasing  Functions 49 

6.  Curve  Tracing 51 

7.  Relative  Maxima  and  Minima.     Points  of  Inflection    .         .         .53 

8.  On  the  Roots  of  Equations 57 

ix 


X  CONTENTS 

CHAPTER  IV 
Differentiation  of  Transcendental  Functions 

ART.  PAGE 

1.  Differentiation  of  sin  a; 62 

2.  Differentiation  of  cos  a;,  tanx,  etc 67 

3.  Inverse  Functions .68 

4.  The  Inverse  Trigonometric  Functions 69 

5.  Logarithms  and  Exponentials       .         .         .        .  .      .         .         .74 

6.  Differentiation  of  log  x 78 

7.  The  Compound  Interest  Law 82 

8.  Differentiation  of  ex,  ax 83 

CHAPTER   V 
Infinitesimals  and  Differentials 

1.  Infinitesimals 85 

2.  Fundamental  Theorem 89 

3.  Tangents  in  Polar  Coordinates .90 

4.  Differentials 91 

5.  Technique  of  Differentiation 94 

6.  Differential  of  Arc 99 

7.  Rates  and  Velocities 101 

CHAPTER  VI 
Integration 

1.  The  Area  under  a  Curve Ill 

2.  The  Integral 114 

3.  Special  Formulas  of  Integration   .        .        .        .        ,        .        .118 

4.  Integration  by  Substitution  .. 119 

5.  Integration  by  Ingenious  Devices 122 

6.  Integration  by  Parts 125 

7.  Use  of  the  Tables 126 

8.  Length  of  the  Arc  of  a  Curve        .        .        .                 .        .        .  129 

CHAPTER  VII 
Curvature.     Evolutes 

1.  Curvature 134 

2.  The  Osculating  Circle 138 


CONTENTS  xi 

ART.  PAGE 

8.   TheEvolute 139 

4.   Properties  of  the  Evolute 143 

CHAPTER  VIII 

The  Cycloid 

1.  The  Equations  of  the  Cycloid 146 

2.  Properties  of  the  Cycloid 147 

3.  The  Epicycloid  and  the  Hypocycloid 149 

CHAPTER   IX 
Definite  Integrals 

1.  A  New  Expression  for  the  Area  under  a  Curve                             .  lj 

2.  The  Fundamental  Theorem  of  the  Integral  Calculus  . 

3.  (Volume  of  a  Solid  of  Revolution 157 

4.  Other  Volumes 159 

5.  Fluid  Pressure 161 

6.  Duhamel's  Theorem 164 

7.  Length  of  a  Curve 166 

8.  Area  of  a  Surface  of  Revolution          ......  167 

9.  Centre  of  Gravity 169 

10.  Centre  of  Gravity  of  Solids  and  Surfaces  of  Revolution      .         .  170 

11.  Centre  of  Gravity  of  Plane  Areas 172 

12.  General  Formulation 174 

13.  Centre  of  Fluid  Pressure 175 

14.  Moment  of  Inertia 176 

15.  A  General  Theorem 179 

16.  The  Attraction  of  Gravitation .181 

17.  Proof  of  Formula  (3).     Variable  Limits  of  Integration               .  184 

CHAPTER   X 
* 

Mechanics 

1.  The  Laws  of  Motion 190 

2.  Absolute  Units  of  Force 194 

3.  Simple  Harmonic  Motion 201 

4.  Motion  under  the  Attraction  of  Gravitation        ....  206 

5.  Constrained  Motion 209 


xii  CONTENTS 

ART.  PAGE 

6.  Motion  in  a  Resisting  Medium 212 

7.  Graph  of  the  Resistance 216 

8.  Motion  under  an  Attractive  Force  with  Damping       .         .         .  218 

9.  Motion  of  a  Projectile 224 


CHAPTER   XI 

The  Law  of  the  Mean.     Indeterminate  Forms 

1.  Rolle's  Theorem  .         . 229 

2.  The  Law  of  the  Mean 230 

3.  Application 231 

4.  Indeterminate  Forms.     The  Limit  § 231 

5.  A  More  General  Form  of  the  Law  of  the  Mean ....  234 

6.  The  Limit  #,  Concluded 235 

7.  The  Limit  = .        .        .        .236 

8.  The  Limit  0  •  oo 238 

9.  The  Limits  0°,  1",  oo°,  and  oo  -  oo 239 


CHAPTER  XII 

Convergence  op  Infinite  Series 

1.  The  Geometric  Series 243 

2.  Definition  of  an  Infinite  Series    ....:..  244 

3.  Tests  for  Convergence 245 

4.  Divergent  Series 248 

5.  The  Test-Ratio  Test 250 

6.  Alternating  Series 252 

7.  Series  of  Positive  and  Negative  Terms ;  General  Case        .        .  254 

8.  Power  Series 257 

9.  Operations  with  Infinite  Series    .                 258 


CHAPTER  XIII 

Taylor's  Theorem 

1.  Maclaurin's  Series .  262 

2.  Taylor's  Series 264 

3.  Proof  of  Taylor's  Theorem 266 

4.  A  Second  Form  for  the  Remainder 269 

5.  Development  of  ex,  sin  x,  cos  x 269 


CONTENTS  xiii 

ART.  PAGE 

6.  The  Binomial  Theorem 272 

7.  Development  of  sin-1  x 274 

8.  Development  of  tanx 274 

9.  Applications 275 

CHAPTER  XIV 

Partial  Differentiation 

1.  Functions  of  Several  Variables.     Limits  and  Continuity   .        .  282 

2.  Formulas  of  Solid  Analytic  Geometry 283 

3.  Partial  Derivatives 287 

4.  Geometric  Interpretation 288 

5.  Derivatives  of  Higher  Order 290 

6.  The  Total  Differential 292 

7.  Continuation.     Change  of  Variable 295 

8.  Conclusion 298 

9.  Euler's  Theorem  for  Homogeneous  Functions    ....  300 

10.  Differentiation  of  Implicit  Functions 302 

11.  A  Question  of  Notation .         .  306 

12.  Small  Errors 306 

13.  Directional  Derivatives 308 

14.  Exact  Differentials 309 

CHAPTER  XV 
Applications  to  the  Geometry  of  Space 

1.  Tangent  Plane  and  Normal  Line  of  a  Surface     ....  316 

2.  Tangent  Line  and  Normal  Plane  of  a  Space  Curve     .        .        .  317 

3.  The  Osculating  Plane 323 

4.  Confocal  Quadrics 326 

5.  Curves  on  the  Sphere,  Cylinder,  and  Cone         ....  329 

6.  Mercator's  Chart 331 

CHAPTER  XVI 

Taylor's  Theorem  for  Functions  of  Several  Variables 

1.  The  Law  of  the  Mean 334 

2.  Taylor's  Theorem 335 

Maxima  and  Minima 336 

4.   Test  by  the  Derivatives  of  the  Second  Order      ....  340 


xiv  CONTENTS 


CHAPTER  XVII 
Envelopes 

ART.  PAGK 

1.  Envelope  of  a  Family  of  Curves 344 

2.  Envelope  of  Tangents  and  Normals 348 

3.  Caustics 350 


CHAPTER  XVIII 

Double  Integrals 

1.  Volume  of  Any  Solid 351 

2.  Two  Expressions  for  the  Volume  under  a  Surface  ;  First  Method  353 

3.  Continuation ;  Second  Method 356 

4.  The  Fundamental  Theorem  of  the  Integral  Calculus  .         .         .  357 

5.  Moments  of  Inertia 359 

6.  Theorems  of  Pappus 362 

7.  Polar  Coordinates 363 

8.  Areas  of  Surfaces 367 

9.  Cylindrical  Surfaces 370 

10.  Analytical  Proof  of  the  Fundamental  Theorem  ;  Cartesian  Co- 

ordinates               ...  371 

11.  Continuation ;  Polar  Coordinates 373 

12.  Surface  Integrals 374 

CHAPTER  XIX 

Triple  Integrals 

1.  Definition  of  the  Triple  Integral 380 

2.  The  Iterated  Integral  '. 382 

3.  Continuation  ;  Polar  Coordinates 386 

4.  Line  Integrals 390 

CHAPTER  XX 
Approximate  Computations.     Hyperbolic  Functions 

1.  The  Problem  of  Numerical  Computation     .        .        .        .        .  398 

2.  Solution  of  Equations.     Known  Graphs 398 

3.  Newton's  Method 399 

4.  Direct  Use  of  the  Tables 402 

5.  Successive  Approximations 403 


CONTENTS  xv 

ART.  PAGE 

6.  Definite  Integrals.     Simpson's  Rule 406 

7.  Amsler's  Planimeter .,        .  409 

8.  The  Hyperbolic  Functions 412 

APPENDIX 

A.  The  Exponential  Function 417 

B.  Functions  without  Derivatives 422 

Supplementary  Exercises 424 

Index         ............  460 


CALCULUS 

CHAPTER   I 

INTRODUCTION 

1.  Functions.  The  student  has  already  met  the  idea  of  the 
function  in  the  graphs  he  has  plotted  and  used  in  Algebra  and 
Analytic  Geometry.     For  example,  if 

the  graph  is  a  straight  line ;  if 

y2  =  2  mx,  y  =  ±  V2mx, 

it  is  a  parabola,  and  if 

?/  =  sin  x, 
we  get  a  succession  of  arches  that  recur  periodically.     Thus  a 
function  was  thought  of  originally  as  an  expression  involving 
x  and  having  a  definite  value  when  any  special  value  is  given 

to  x: 

f(x)=2x  +  3, 


or  f(x)  =  ±  V2ma?, 

or  f(x)  =  sin  x. 

Other  letters  used  to  denote  a  function  are  <j>(x),  \f/(x)}  F(x), 
etc.     We  read /(a;)  as  "/of  xP 

Further  examples  of  functions  are  the  following:  (a)  the 
volume  of  a  sphere,  V,  as  a  function  of  the  length  of  the 
radius,  r : 

r-.4«f«i 


2  CALCULUS  , 

(&)  the  distance  s  that  a  stone  falls  when  dropped  from  rest,  as 
a  function  of  the  time  t  that  it  has  been  falling : 

(c)  the  sum  of  the  first  n  terms  of  a  geometric  progression : 

sn=  a  +  ar  -f-  ar2  H -f  arn~l, 

as  a  function  of  n : 

a  —  arn 


s„  = 


1-r 


In  higher  mathematics  the  conception  of  the  function  is 
enlarged  so  as  to  include  not  merely  the  case  that  y  is  actually 
expressed  in  terms  of  x  by  a  mathematical  formula,  but  also 
the  case  of  any  law  whatever  by  which,  when  x  is  given,  y 
is  determined.  We  will  state  this  conception  as  a  formal 
definition. 

Definition  of  a  Function,  y  is  said  to  be  a  function  of  x 
if  when  x  is  given,  y  is  determined. 

As  an  example  of  this  broader  notion  of  the  function  con- 
sider the  curve  traced  out  by  the  pen  of  a  self -registering  ther- 
mometer. Here,  a  sheet  of  paper  is  wound  round  a  drum 
which  turns  slowly  and  at  uniform  speed,  its  axis  being  ver- 
tical. A  pen,  pressing  against  this  drum,  is  attached  to  a 
thermometer  and  can  move  vertically  up  and  down.  The 
height  of  the  pen  above  the  lower  edge  of  the  paper  depends 
on  the  temperature  and  is  proportional  to  the  height  of  the 
temperature  above  a  given  degree,  say  the  freezing  point. 
Thus  if  the  drum  makes  one  revolution  a  day,  the  curve  will 
show  the  temperature  at  any  time  of  the  day  in  question.  The 
sheet  of  paper,  unwound  and  spread  out  flat,  exhibits,  then,  the 
temperature  y  as  a  function  of  the  time  x* 

*  Objection  has  been  raised  to  such  illustrations  as  the  above  on  the 
ground  that  the  ink  mark  does  not  define  y  accurately  for  a  given  x,  since 
the  material  graph  has  appreciable  breadth.  True  ;  but  we  may  proceed 
here  as  in  geometry,  when  we  idealize  the  right  line.  What  we  see  with 
our  eyes  is  a  taut  string  or  a  line  drawn  with  a  ruler  or  a  portion  of  the 


INTRODUCTION  3 

Other  examples  of  this  broader  conception  of  the  function 
are :  the  pressure  per  square  inch  of  a  gas  enclosed  in  a  vessel, 
regarded  as  a  function  of  the  temperature ;  and  the  resistance 
of  the  atmosphere  to  the  motion  of  a  rifle  bullet,  regarded  as  a 
function  of  the  velocity. 

A  function  may  involve  one  or  more  constants,  as,  for 
example : 

f(x)=ax-\-b,  <f>  (x)  =  tan  ax. 

Here,  a  and  b  are  any  two  numbers,  which,  however,-  once 
chosen,  are  held  fast  and  do  not  vary  with  x. 

If  y  is  a  function  of  x,  y  =f(x),  then  x  is  called  the  indepen- 
dent variable  and  y  the  dependent  variable.  The  independent 
variable  is  the  one  which  we  think  of  as  chosen  arbitrarily, 
i.e.  we  assign  to  it  at  pleasure  any  values  which  it  can  take 
on  under  the  conditions  of  the  problem.  The  other  variable 
or  variables  are  then  determined.     Thus  when  we  write: 

s=16f, 

we  think  of  t  as  the  independent,  s  as  the  dependent  variable. 
But  if  we  solve  for  t  and  write : 

l~  4' 

then  we  think  of  s,  which  is  here  necessarily  restricted  to  posi- 
tive values,  as  the  independent,  t  as  the  dependent  variable. 
In  general,  if  two  variables  are  connected  by  a  single  equation, 

as  for  example  ~ 

pv  =  C, 

where  C  is  a  constant,  either  may  be  chosen  as  the  indepen- 
dent variable,  the  other  thus  becoming  the  function. 

path  of  a  stone  thro  ?n  hard.  These  are  not  straight  lines  ;  but  they  sug- 
gest a  concept  obtained  by  thinking  of  finer  and  ever  finer  threads  and 
narrower  and  ever  narrower  lines,  and  thus  we  get  at  the  straight  lines  of 
geometry.  So  here,  we  may  think  of  the  actual  temperature  at  each 
instant  as  having  a  single  definite  value  and  thus  being  a  function  of  the 
time,  the  ideal  graph  of  this  function,  then,  being  a  geometric  curve  that 
lies  within  the  material  belt  of  ink  traced  out  by  the  pen. 


4  CALCULUS 

A  function  may  depend  on  more  than  one  independent  vari- 
able. Thus  the  area  of  a  rectangle  is  equal  to  the  product  of 
two  adjacent  sides.     Further  examples: 

/0,2/)  =  ilog  02  +  2/2)> 

<f>  (x,  y,  z)  =  ax2  +  by2  -f-  cz2  —  d. 

Again,  it  may  happen  that  to  one  value  of  x  correspond 
several  values  of  y,  as  when 


x2  +  y2  =  a2,  y=z±va2  —  x2. 

y  is  then  said  to  be  a  multiple-valued  function  of  x.  But  usually, 
when  this  is  the  case,  it  is  natural  to  group  the  values  so  as 
to  form  a  number  of  single-valued  functions.  In  the  above 
example  these  will  be* 


y  =  Va2  —  x2        and        V  —  —  Va2  —  x2. 

In  this  book  a  function  will  be  understood  to  be  single- 
valued  unless  the  contrary  is  explicitly  stated. 

A  function  is  said  to  be  continuous  if  a  small  change  in  the 
variable  gives  rise  to  a  small  change  in  the  function.  Thus 
the  function 

1 

y  =  -> 

X 

whose  graph  is  a  hyperbola,  is  in  general  continuous  ;  but 
when  x  approaches  0,  y  increases  numerically  without  limit, 
and  so  at  the  point  x  —  0  the  function  is  discontinuous. 

*  The  sign  y/  means  the  positive  square  root  of  the  radicand,  not, 
either  the  positive  or  the  negative  square  root  at  pleasure.  Thus,  V4  is 
2,  and  not  —  2.  This  does  not  mean  that  4  has  only  one  square  root. 
It  means  that  the  notation  VI  calls  for  the  positive,  and  not  for  the 
negative,  of  these  two  roots.     Again, 

Vz2  =   x,    if  a;  is  positive ; 
VaJ2  =  _  x,  if  x  is  negative. 

A  similar  remark  applies  to  the  symbol  2^/,  which  likewise  is  used  to 
mean  the  positive  2  wth  root. 


INTRODUCTION 


It  is  frequently  desirable  to  use  merely  the  numerical  or 
absolute  value  of  a  quantity,  and  to  have  a  notation  for  the 
same.  The  notation  is :  \x\,  read  "  the  absolute  value  of 
x."  Thus,  |  —  3  |  =  3  ;  1 3  |  =  3.  Again,  whether  a  be  posi- 
tive or  negative,  we  always  have 

Va2  =  |  a\. 

2.  Slope  of  a  Curve.  We  have  learned  in  Analytic  Geometry 
how  to  find  the  slope  of  some  of  the  simpler  curves.  The 
method  is  of  fundamental  importance  in  the  Calculus,  and  so 
we  begin  by  recalling  it. 

Consider,  for  example,  the  parabola  : 


(1) 


y 


X2. 


Let  P,  with  the  coordinates 
(x0,  2/0),  be  an  arbitrary  point 
on  this  curve,  and  let 
P':  (x',  y'),  be  a  second  point. 
Pass  a  secant  through  P  and 
F.     Let 


x'  =  x0  +  h, 


2/o  +  &• 


Then  the  slope  of  the  secant 
will  be : 

tanr'  = 


O 
Fig.  1 


.'/o 


a% 


When-Pf  approaches  P  as  its  limit,  the  secant  rotates  about  P 
and  approaches  the  tangent  as  its  limit,  its  slope  approaching 
the  slope  of  the  tangent :  * 

lim  taiiT'  =  taiiT. 
p=p 

We  wish  to  evaluate  this  limit. 

*  The  sign.=  means  that  limit,  without,  how- 

ever, ever  being  allowed  to  reach  this  limit.     For,  if  i*  were  to  coincide 
with  P,  we  should  no  longer  have  a  determinate  secant,  one  point  not 
to  determine  a  straight  line. 


6 


CALCULUS 


Suppose  the  point  Pis  the  point  (1,  1).  Let  us  compute  k 
and  tan  r'  =  k/h  for  a  few  values  of  h.  Here,  x0  =  1,  yQ  =  1. 
Ifh=.l,  I  , 

then  x1  —  x0  +  h  =  1.1,        y1  =zy0  +  k  =  1.21 

and  k  =  y'-y0  =  .21; 


hence 


tanr'  =  -  =  2.1. 


The  following  table  shows  further  sets  of  values  of  h,  k,  and 
tan  r'  that  belong  together : 


ft 

A". 

tan  r  =  j 

.1 

.01 

.001 

.21 
.0201 

.002001 

2.1 

2.01 
2.001 

It  is  the  last  column  that  we  are  chiefly  interested  in,  and 
its  numbers  appear  to  be  approaching  nearer  and  nearer  to  2 
as  their  limit.  Let  us  prove  that  this  is  really  so.  As  the  proof 
is  just  as  simple  for  an  arbitrary  point  P,  we  will  return  to  the 
general  case. 

Since  P  and  P'  lie  on  the  curve  (1),  we  have  : 

(2)  y0  =  x02, 

(3)  yQ  +  k  =  (x0  +  hy  =  x*  +  2xQh  +  h\ 
Hence,  subtracting  (2)  from  (3),  we  get : 

(4)  k  =  2x0h  +  h2, 

k 

it 


and  finally : 


tanr' 


ixo  +  h. 


Now  let  P'  approach  P  as  its  limit.     Then  h  approaches  0  and 
we  have : 

k 
lim  tan  r'  =  lim-  =  lim(2a?0  4-  7i), 

P'=P  7i=0  Jl  h±Q 

(5)  tanr  =  2ic0. 

If,  in  particular,  sc0=l,  then  tan  t=2,  as  we  set  oux  to  prove. 


INTRODUCTION  7 

EXERCISES 

1.    If  f(x)  =  x*-3x  +  2, 

show  that  /(l)  ==  0,  and  compute  /(0),  /(-  1),  /(1£). 

find/(V2)  correct  to  three  significant  figures.       Ans.  —.0204. 

3.  If  F(x)  =  (x  —  x3)smx, 
find  all  the  values  of  x  for  which  F(x)  =  0. 

4.  If  a     <K*)  =  2S, 
find  *(<&+(- 3),  *(*). 


5.  If  /(a>)=»-Va*-a>", 

find  "/(a) and  /(0). 

5-7T 

6.  If  in  the  preceding  question  a  =  cos  — -,  compute /(0)  to 

6 

three'  significant  figures. 

7.  If  ij/(x)  =  x$  —  x~%, 
find  if/(S). 

8.  If  f(x)  =  x\og10(12-x*), 
find/(-2)and/(3i). 

9.  Solve  the  equation 

Xs  —  xy  -\-  S  =  5y 
for  y,  thus  expressing  y  as  a  function  of  x, 

10.  If  A*)838*! 
show  that                   /(a?)  /(?/) « A*  +  2/)- 

11.  Continue  the  table  of  §  2  two  lines  further,  using  the 
values  h  =  .0001  and  h  =  .00001. 


8  CALCULUS 

12.  Find  the  slope  of  the  curve 

Sy  =  Sxs 

at  the  point  (2,  3),  first  preparing  a  table  similar  to  that  of  §  2 
and  then  proving  that  the  apparent  limit  is  actually  the  limit. 

13.  Find  the  slope  of  the  curve 

y  =  Xs  —  x2 
at  any  point  (x0,  yQ). 

14.  Find  the  slope  of  the  curve 

y  =  ax2  +  bx  +  c 
at  the  point  (x0,  y0). 

15.  Find  the  slope  of  the  curve 


X 


at  Oo,  2/o). 


CHAPTER  II 

DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS 
GENERAL   THEOREMS 

1.  Definition  of  the  Derivative.  The  Calculus  deals  with 
varying  quantity.  If  y  is  a  function  of  x,  then  x  is  thought 
of,  not  as  having  one  or  another  special  value,  but  as  flowing 
or  growing,  just  as  we  think  of  time  or  of  the  expanding  cir- 
cular ripples  made  by  a  stone  dropped  into  a  placid  pond. 
And  y  varies  with  x,  sometimes  increasing,  sometimes  de- 
creasing. Now  if  we  consider  the  change  in  x  for  a  short 
interval,  say  from  x  =  x0  to  x  =  x',  the  corresponding  change 
in  y,  as  y  goes  from  y0  to  y\  will  be  in  general  almost  pro- 
portional to  the  change  in  x,  as  we  see  by  looking  at  the 
graph  of  the  function ;  for 

X      O/Q 

and  tanr'  approaches  the  limit  tanr,  i.e.  comes  nearer  and 
nearer  to  the  fixed  value  tan  t,  the  slope  of  the  tangent. 
The  determination  of  this  limit : 

lim  y^^  =  ttmT, 

z'±xo  X   —  Xq 

is  a  problem  of  first  importance,  and  we  shall  devote  the  next 
few  chapters  to  solving  it  for  the  functions  we  are  already 
familiar  with  and  to  giving  various  applications  of  the 
results. 

9 


10  CALCULUS 

We  will  first  formulate  the  idea  we  have  just  explained  as  a 
definition.     Let 

(i)  y=m 

be  a  given  function  of  x.     Form  the  function  for  an  arbitrary- 
value  x0  of  x : 

(2)  2/o  =/(*«), 

and  then  give  to  x0  an  increment,   Ax-,  i.e.  consider  a  second 
value  x'  of  x  and  denote  the  difference  x'  —  x0  by  the  symbol  * 

Ax: 

x'  —  x0  =  Ax  ;  x'  =  x0  +  Ax. 

The  function  ?/  will  thereby  have  changed  from  the  value  ya  to 
the  value 

(3)  V'  =/(*») 

and  hence  have  received  an  increment 

y'-y0==Ay;  \j  =  y0  +  Ay. 

From  (3),  written  in  the  form : 

(4)  y0  +  Ay=f(x0  +  Ax), 

and  (2),  we  obtain  by  subtraction : 

Ay  =f(x0  +  Ax)  -f(x0), 
L  hence 


and  hence 

A.y =/Oo  +  Aa0  -ZOO 

Ax  Ax 


Definition  of  a  Derivative.  The  limit  which  the  ratio  (5) 
approaches  ivhen  Ax  approaches  0 : 

*  The  student  must  not  think  of  this  symbol  as  meaning  A  times  x. 
We  might  have  used  a  single  letter,  as  h,  to  represent  the  difference  in 
question  :  x'  =  x0  +  h  ;  but  h  would  not  have  reminded  us  that  it  is  the 
increment  of  #,  and  not  of  y,  with  which  we  are  concerned.  The  nota- 
tion is  read  "  delta  aj." 


DIFFERENTIATION'  OF   ALGEBRAIC   FUNCTIONS       11 


(6) 


Km  £*       or       lim  /(*•  +  **)-/(*.) 

Ax  =  0A.X"  Ax  =  0  AX 


is  called  the  derivative  of  y  with  respect  to  x  and  is  denoted  by 
Dxy,  (read:  "Dxofy"): 


(?) 


As=o  Ax 


Dxy. 


y 

pJ// 

m 

J^L 

~'\ 

Ay 

^-~-_____— ~<S^  !& 

\ 

1 

i 

k 

i 
i 

a; 

0 

•% 

a 

Fig.  2 


In  the  above  definition  Ax  may  be  negative  as  well  as  posi- 
tive and  the  limit  (6)  must  have  the  same  value*  when  Ax 
approaches  0  from  the 
negative  side  as  when 
it  approaches  0  from 
the  positive  side. 

Our  problem  is,  then, 
to  compute  the  deriva- 
tive for  the  various 
functions  that  present 
themselves. 

To  differentiate  a 
function  is  to  find  its 
derivative. 

The  geometrical  interpretation  of  the  analytical  process  of 
differentiation  is  to  find  the  slope  of  the  graph  of  the  function. 

2.   Differentiation  of  a**.     Let 

(8)  '  ?  =  af, 

where  n  is  a  positive  integer.  To  differentiate  y  we  are  first 
to  give  x  an  arbitrary  value  x0  and  compute  the  corresponding 
value  y0  of  the  function  : 

(9)  2/o  =  *on. 

Next,  give  to  x  an  increment,  Ax,  whereby  y  receives  a  corre- 
sponding increment,  Ay : 

(10)  y0  +  Ay  =  (x0  +  Axy. 


12  CALCULUS 

Expanding  the  right-hand  member  by  the  Binomial  Theorem, 
we  get : 

(11)  2/0  +  Ay  =  xf  +  nx^Ax  +  n^~1)  x0n-2Ax2  +  •••  4-  Aaf . 

If  we  subtract  (9)  from  (11)  and  divide  through  by  Ax,  we  get : 

^  =  nxf-1  +  n^~1)  x,n~2Ax  +  •  •  •  +  Aa""1. 
Ax  1.2 

We  are  now  ready  to  allow  A#  to  approach  0  as  its  limit. 
On  the  right-hand  side  the  first  term  is  constant.  Each  of  the 
succeeding  terms  approaches  0  as  its  limit,  and  so  their  sum 
approaches  0.     Hence  we  have  : 

iim  — -  =  nxQn~l. 

Az=0  AX 

The  subscript  of  the  x0  has  now  served  its  purpose  of 
reminding  us  that  x0  is  not  to  vary  with  Ax.  In  the  final 
result  we  may  drop  the  subscript,  for  x0  is  any  value  of  x,  and 
thus  we  obtain  the  formula,  or  theorem  : 

(12)  Dxxn  =  nxn~\ 
In  particular,  if  n  =  1,  we  have 

(13)  Dxx  =  l. 

EXERCISES 

1.  Show  that 

Dx(cxn)  =  ncxn-\ 
where  c  is  a  constant. 

2.  Write  down  the  derivatives  of  the  following  functions : 

x2,      a?,       x49,       7x,       -9x\ 

3.  Differentiate  the  function : 

V  =  i.  Arts.      -\- 

y     x2  a? 

4.  Differentiate:  u  —  t2  —  t. 


»IFFERENTIATI@N  #F   ALGEBRAIC   FUNCTIONS      13 

3.  Derivative  of  a  Constant.  If  the  function  f(x)  is  a 
constant : 

the  graph  is  a  straight  line  parallel  to  the  axis  of  x,  and  so  the 

slope  is  0.     Hence 

(14)  Dxc  =  0. 

It  is  instructive  to  obtain  this  result  analytically  from  the 
definition  of  §  1.     We  have: 

y0  +  Ay  =/Oo  +  Aa;)  =  c, 

hence  Av=0        and        — ^  =  0. 

a  Ax 

Allowing  Ax  now  to  approach  0  as  its  limit,  we  see  that  the 
value  of  the  variable,  Ay /Ax,  is  always  0,  and  hence  its  limit 
isO: 

lim^/  =  0         or       Dxc  =  0. 

As=oAaj 

4.  General  Formulas  of  Differentiation. 

Theorem  I.  The  derivative  of  the  product  of  a  constant  and 
a  function  is  equal  to  the  product  of  the  constant  into  the  deriva- 
tive of  the  function : 


(I) 

Dx(cu)  =  cDxu. 

For,  let 

y=cu. 

Then 

y<>  =  cu0, 

y0  +  Ay  =  c(u0  +  Au)i 

hence 

Ay  =  cAu, 

Ax        Ax' 

and 

lira  *M=  lira  (c^\ 

Az=oA#      Ax=oV    Ax) 

14  CALCULUS 

The  limit  of  the  left-hand  side  is  Dxy.  On  the  right.  An/ 'Ax 
approaches  Dxu  as  its  limit,  hence  the  limit  of  the  right-hand 
side  is*  cDxu,  and  we  have 

Dx  (cu)  —  cDx  u,  q.  e.  d. 

Theorem  II.    The  derivative  of  the  sum  of  two  functions  is 
equal  to  the  sum  of  their  derivatives : 


(II) 

Dx(u  +  v)  =  Dxu  +  Dxv. 

For, 

let 

y  =  u  +  v. 

Then 

y0  =  u0  +  v0} 
y0  +  Ay  =  Uq  +  Au  +  %  +  Av, 

hence 

Ay  =  Au  +  Av, 

and 

Ay  _  Au      Av 

Ax      Ax      Ax 

When  Ax  approaches  0,  the  first  term  on  the  right  approaches 
Dxu  and  the  second  Dxv.  Hence  the  whole  right-hand  side 
approaches  *  Dxu  -+-  Dxv,  and  we  have 


t      Aw      v      [Au  ,  Av\      t      Au  .  v      &v 

lim  — £  ==  lira  [ 1 \  =  hm f-  hm  — 

Ax=oAa?      Ax=o^Aa7      Ax)      Ax=o  Ax      Ax=oAx 


or  Dxy  =  Dxu  +  Dxv,  q.  e.  d. 

Corollary.     The  derivative  of  the  sum  of  any  number  of 
functions  is  equal  to  the  sum  of  their  derivatives. 

If  we  have  the  sum  of  three  functions,  we  can  write 

U-\-V  +  W=:U+(v-\-  W). 

Hence  Dx  (u  +  v  +  w)  =  Dxu  +  Dx  (v  -f  w) 

=  Dxu  +  Dxv  +  Dxw. 

*  For  a  careful  proof  of  this  point  cf.  §  5. 


DIFFERENTIATION  OF   ALGEBRAIC   FUNCTIONS      15 

Next,  we  can  consider  the  snm  of  four  functions,  and  so  on. 
Or  we  can  extend  the  proof  of  Theorem  II  immediately  to  the 
sum  of  n  functions. 

Polynomials.  We  are  now  in  a  position  to  differentiate  any 
polynomial.     For  example : 

D„(7a?*-5<e»+a?  +  2) 

=  7Dxx*-5Dxx*  +  1  =  28^-15^  +  1. 

EXERCISES 

Differentiate  the  following  functions : 

1.   5^-8^  +  7^-^  +  1.  3.   Trar3- 41^-^/2. 

8  a7 -6a; +  5  .  ax2  +  2hx+b 

9  2c 

5.  Differentiate 

(a)  v0t  — 16 t2  with  respect  to  t. 

(b)  a  +  bs  +  cs2  with  respect  to  s. 

(c)  .  01  ly*  —  8.15  my2  — .  9  Im  with  respect  to  y. 

6.  Find  the  slope  of  the  curve 

4:y  =  x*  —  Sx  —  1 
at  the  point  (1,  —  2). 

7.  At  what  angles  do  the  curves  y=x*  and  y=x*  intersect? 

Ans.   0°  and  8°  V. 
/  5.   Three  Theorems  about  Limits. 

Theorem  A.  TJie  limit  of  the  sum  of  two  variables  is  equal 
to  the  sum  of  their  limits : 

lim  (X+  Y)  =  lim  X  +  lim  T. 


16  CALCULUS 

Let  limX=^4,  lim  Y=B, 

and  let  c  denote  the  difference  between  X  and  its  limit,  A : 
X-A  =  e,  X=A  +  e. 

Then  lim  e  =  0. 

(If  X  is  less  than  A,  c  will  be  a  negative  quantity.) 
In  like  manner  let 

Y-B  =  7],  Y=B  +  V. 

Then  lim^  =  0. 

Writing  now 

X+Y=A  +  B  +  €  +  n; 

we  see  that  the  limit  of  the  right-hand  side  is  A  -f  B.     Hence 

lim  (X  +  Y)=A  +  B,  q.  e.  d. 

Corollary.      TJie  limit  of  the  sum  of  n  variables  is  equal  to 
the  sum  of  the  limits  of  these  variables,  n  being  any  fixed  number : 

lim  (Xx  +  X2  +  ...  +  Xn)  =  lim  Xx  +  lim  X>  +  •  •  •  +  lim  Xn.  ' 

Theorem  B.     TJie  limit  of  a  product  is  equal  to  the  product 
of  the  limits: 

lim  (IF)=(lim  X)(lim  Y).  * 

Here  XY  =  (A  +  e)  (B  +  v) 

=  AB  +  Arj  +  Be  +  erj. 

By  Theorem  A,  Corollary,  the  limit  of  the  right-hand  side  is 

AB  +  lim  (Av)  +  lim  (Be)  +  lim  (erj). 

The  last  limit  is  obviously  0.  As  regards  the  first  two,  it  is 
easy  to  see  that  *  if  a  variable  (as  rj)  approaches  0,  then  the 
product  of  any  constant  (as  A)  times  this  variable  must  also 
approach  0.  An  unfavorable  case  would  be  that  in  which  the 
constant  is  very  large,  say  10,000,000.  But  even  then  the 
variable,  as  it  decreases,  will  finally  become  and  remain 
numerically  less  than  10"7  =  100J)(X)0,  and  so  the  product  becomes 


DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS      17 

less  than  1.  As  the  variable  decreases  still  further,  it  becomes 
and  remains  numerically  less  than  10~8,  then  less  than  10~9, 
and  so  on ;  the  product  thus  becoming  and  remaining  numeri- 
cally less  than  y1^,  y-J-^,  and  so  on.  Hence  the  limit  of  the 
product  is  0. 

Thus  each  of  the  limits  in  the  above  expression  is  seen  to  be 
0,  and  hence  lim(XY)=AB,  q.e.d. 

In  particular,  we  see  that  the  limit  of  a  constant  times  a  vari- 
able is  equal  to  the  product  of  the  constant  into  the  limit  of  the 
variable:  lim((7X)  =  Clim(X). 

For,  a  constant  is  a  special  case  of  a  variable. 

Corollary.  The  limit  of  the  product  of  n  variables  is  equal 
to  the  product  of  their  limits,  n  being  any  fixed  number  : 

lim  (X,  X2  •  •  •  Xn)  =  (lim  Xx)  (lim  X2)  ■ . .  (lim  Xn) . 

Theorem  C.  The  limit  of  the  quotient  of  two  variables  is 
equal  to  the  quotient  of  their  limits,  provided  that  the  limit  of  the 
variable  that  forms  the  denominator  is  not  0 : 

Hffi?=-iE-|L         if         limF^O. 

Y  lim  Y 

For  X      A-A  +  *     A=Be-Av 

Y  B     B  +  n     B     B(B  +  V)' 

X_A     Be-Av       1 
Y~B+       &  £ 

B 

The  limit  of  the  first  fraction  in  the  last  term  is  0,  by  Theo- 
rems A  and  B.  The  second  fraction  ultimately  becomes  posi- 
tive and  remains  less  than  2,  even  if  rj  is  negative.  For,  since 
lim  77  =  0,     /B  will  finally  become  and  remain  algebraically 


greater  than  - 


1  . 


V  ^-1    .   V 


1+J 


18  CALCULUS 

Hence  the  last  term  becomes  and  remains  numerically  less  than 
twice  the  first  factor,  and  consequently  its  limit  is  0. 

•••limf  =  |,  1-e.d. 

In  particular,  we  see  that,  if  a  variable  approaches  unity  as 
its  limit,  its  reciprocal  also  approaches  unity  : 

If    limX=l,         then         limi  =  l. 

Remark.  If  the  denominator  Y  approaches  0  as  its  limit, 
no  general  inference  about  the  limit  of  the  fraction  can  be 
drawn,  as  the  following  examples  show.  Let  Y  have  the 
values : 

Y=i_  j_  __i_       _i_ 

10'  100'  1000'  '  "'  10"'  " 
(1)  If  the  corresponding  values  of  X  are  : 

x=l_  _i L_       J_ 

102'  1002'  10002'  '"'  102n'  '", 
then  lim  —  =  lim  —  =  0. 


(2)     If         X  = 


Y  10» 

111  1 


Vio'  yioo'  yiooo'    '  1Q« 


then  X/Y=10n/2  approaches  no  limit,  but  increases  beyond 
all  limit. 

/Q\   Tf  Y"  —   c         c  c        . . .       G 

V)  ±-1Q>  100>  1000>      .  1Qn» 

where  c  is  any  arbitrarily  chosen  fixed  number,  then 

lim  |= ft 


DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS      19 
111  1 


(4)  If  X  = 


10'       100'  1000'       10,000' 


then  X/Y  assumes  alternately  the  values  +1  and  —1,  and 
hence,  although  remaining  finite,  approaches  no  limit. 

To  sum  up,  then,  we  see  that  when  X  and  Y  both  approach 
0  as  their  limit,  their  ratio  may  approach  any  limit  whatever, 
or  it  may  increase  beyond  all  limit,  or  finally,  although  remain- 
ing finite,  i.e.  always  lying  between  two  fixed  numbers,  no  mat- 
ter how  widely  the  latter  may  differ  from  each  other  in  value, 
—  it  may  jump  about  and  so  fail  to  approach  a  limit. 

Infinity.  If  lim  X=A=f=0  and  lim  Y=0,  then  X/Y  in- 
creases beyond  all  limit,  or  becomes  infinite.  A  variable  Z  is 
said  to  become  infinite  when  it  ultimately  becomes  and  remains 
greater  numerically  than  any  preassigned  quantity,  however 
large.  If  it  takes  on  only  positive  values,  it  becomes  positively 
infinite;  if  only  negative  values,  it  becomes  negatively  infinite. 
We  express  its  behavior  by  the  notation : 

limZ=oo      or     limZ=-fco      or     \\mZ  =  —  oo . 

But  this  notation  does  not  imply  that  infinity  is  a  limit ;  the 
variable  in  this  case  approaches  no  limit.  And  so  the  nota- 
tion should  not  be  read  "  Z  approaches  infinity  "  or  "  Z  equals 
infinity  ;  "  but  "  Z  becomes  infinite." 

Thus  if  the  graph  of  a  function  has  its  tangent  at  a  certain 
point  parallel  to  the  axis  of  ordinates,  we  shall  have  for  that 
point : 

lim  — *  =  oo  ; 

Ax  =  0  AX 

read:    "  Ay /Ax  becomes  infinite  when  Ax  approaches  0." 

Some  writers  find  it  convenient  to  use  the  expression  "a 
variable  approaches  a  limit"  to  include  the  case  that  the  vari- 
able becomes  infinite.  We  shall  not  adopt  this  mode  of  ex- 
pression, but  shall  understand  the  words  "  approaches  a  limit  " 
in  their  strict  sense. 


20  CALCULUS 

If  a  function  f(x)  becomes  infinite  when  x  approaches  a  cer- 
tain value  a,  as  for  example 

f(x)  =  -     for     a  =  0, 
x 
we  denote  this  by  writing 

/(a) =00 

(or    /(a)  =  -f-  co      or     =  —  co ,     if  this  happens  to  be  the  case 
and  we  wish  to  call  attention  to  the  fact). 

Definition  of  a  Continuous  Function.  We  can  now  make  more 
explicit  the  definition  given  in  Chapter  I  by  saying :  f(x)  is 
continuous  at  the  point  x  =  a  if 

lim  f(x)=f  (a). 

From  Exercises  1-3  below  it  follows  that  the  polynomials 
are  continuous  for  all  values  of  x,  and  that  the  fractional 
rational  functions  are  continuous  except  when  the  denominator 
vanishes. 

EXERCISES 

1.  Show  that,  if  n  is  any  positive  integer, 

lim(Xw)  =  (limX)rt. 

2.  If  O (x)  =  c0  +  cxx  +  c2x2  H \-  cnxn, 

then  lim O (x)  =  G(a)  =  c0  4  c^a 4-  c2a? -f  •••  4-  cnan: 

x=a 

3.  If  G  (x)  and  F(x)  are  any  two  polynomials  and  if  F(a)  =£  0, 

then  lim^M  =  ^M. 

«*•  F(x)      F(a) 

4.  If  X  remains  finite  and  Y approaches  0  as  its  limit,  show 

that 

lim(XY)  =  0. 

5.  Show  that 

,.  x2  +  l  1 

lim =  -  • 

*=oo3x2  +  2a;-l     3 


DIFFERENTIATION  OF  ALGEBRAIC   FUNCTIONS      21 

Suggestion :      Begin   by  dividing   the   numerator   and   the 
denominator  by  x2. 

6.    Evaluate  the  following  limits : 

/  \      v          x  + 1                               /  jn     v     ax  -f-  bx~x 
(a)      lim-— — -;  (d)     hm — — -; 


(b)     hm  — - — — - — ■ — -;  (e)     lim  — — — — ; 

(C)     limflag+6flr);  (/)     lim^=* 

w     x=:oo  ex -f  dec"1  ^Vl  +  a4        *. 

6.   General  Formulas  of  Differentiation,  Concluded. 

Theorem  III.     The  derivative  of  a  product  is  given  by  the 
formula : 
(III)  Dx(uv)  =  uDxv  +  uDx?*. 

Let  2/  =  uv. 

Then  y0  =  «o% 

y0  +  Ay  =  (uq  +  Aw)  O0  +  Ay), 

Ay  =  w0  Av  -f-  v0Aw  -f-  Au  Av, 

Aw  Av  .       Aw   ,    A    Av 

^  =  M         +  v         +  Am-—, 

Ax  Ax         Ax  Ax 

and,  by  Theorem  A,  §  5 : 

lim  ^=  lira  L*?)  +  lim  (V-")  +  lim  (Au^Y 

a*=oAx      Ax=o\^    Aay      Axio\^    A.ty       ax^o  \      Aay 

By  Theorem  B,  §  5,  the  last  limit  has  the  value  0,  since 
lim  Au  =  0  and  lim  (Av/ Ax)  =  Dxv.  The  first  two  limits  have 
the  values  u0Dxv  and  v0Dxu  respectively.*  Hence,  dropping 
the  subscripts,  we  have : 

Dxy=  uDxv  +  vDx  u,  q.  e.  d. 

*  More  strictly,  the  notation  should  read  here,  before  the  subscripts  are 
dropped:  [Dxv']x=x  ,  etc.  Similarly  in  the  proofs  of  Theorems  I,  II, 
and  V. 


22  CALCULUS 

By  a  repeated  application  of  this  theorem  the  product  of  any 
number  of  functions  can  be  differentiated.     When  more  than 
two  factors  are  present,  the  formula  is  conveniently  written  in 
the  form : 
(15)  Dx(uvw)  =  Dxu  [  Dxv  j  Dxw 

UVW  U  V  w 

For  a  reason  that  will  appear  later,  this  is  called  the  loga- 
rithmic derivative  of  uvw. 

Theorem  IV.     The  derivative  of  a  quotient  is  given  by  the 
formula:*  ,.      vD  D 

<ro  D\v)=-    *      ' 


Let  y  =  -  • 

v 

mi  ui)  i     a  UQ  +  ^M 

Then  Vo=  —  f  y0  +  AM  =  -°^A    , 

_  ^o  +  Am  _  Mo  __  ^0Am  — MqAv 
V0  +  Av      v0        v0(vo  +  Av)' 

Am  Av 

Vo— -  Mo- 
AM  _      &'x         Aa; 

A«~"  VoOo  +  Av) 
By  Theorem  C  of  §  5  we  have : 


Art 


HmU— 


M, 


"0 

Am_Ax-°\     A# 
Ax=oAx~~  lim   [v0(v0  +  Av)] 

Ax  =  0 

Applying  Theorems  A  and  B  of  §   5  and  dropping  the  sub- 
scripts we  obtain : 

D  Saj^J&t,  q.e.d. 

*  The  student  may  find  it  convenient  to  remember  this  formula  by- 
putting  it  into  words:  "The  denominator  into  the  derivative  of  the 
numerator,  minus  the  numerator  into  the  derivative  of  the  denominator, 
over  the  square  of  the  denominator." 


DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS      23 

Example.     Let 

y  = ! — ■• 

cx  +  d 

Then  D     —  (ca?  +  d)  a  —  {ax  -f-  6)  c  _  ad  — be  # 

Xl/~  (cx  +  d)2  "(caj  +  d)2' 

Theorem  V.  If  u  is  expressed  as  a  function  of  y  and  y  in 
turn  as  a  function  of  x : 

u=f(y),  y  =  <t>(x), 

then 

(V)  ft/(y)  =  ft/(y)ftjr 

or 

-    (V)  Dxu  =  Dyu.Dxy. 

Here         y0  =  <£  (x0),  uQ  =/(y0), 

y0  +  Ay  =  <£  (xq  +  Ax),  u0  +  Au  =/(y0  +  Ay), 

Att=/Gfo  +  %)-/(y0), 

A^ _ f(y0  +  Ay)  — /(y0)  <  Ay# 
Ax  Ay  Ax 

When  Ax  approaches  0,  Ay  also  approaches  0,  and  hence  the 
limit  of  the  right-hand  side  is 

( lim  fbh+m) -/<*)) (nm  *i) = D  f(y) DxV. 

\a*=o  Ay  /\Ax=oAxy 

The  limit  of  the  left-hand  side  is  Dxu,  and  hence 

Dxu  =  Dyu-Dxy,  q.  e.  d. 

The  truth  of  the  theorem  does  not  depend  on  the  particular 
letters  by  which  the  variables  are  denoted.  We  may  replace, 
for  example,  x  by  t  and  y  by  x.  Dividing  through  by  the  sec- 
ond factor  on  the  right,  we  thus  obtain  the  formula : 

tv<)  ft.,,  fis. 


24  CALCULUS 

Example.     Let  u  =  (ax  -f-  6)n, 

where  n  is  a  positive  integer. 

Set  y  =  ax-\-b.  Then  u=f(y)  =  yn 

and       Dx «  =  Dx2/n  —  Dyyn  -  Bxy  —  nyn~x  •  a  =  wa(ax  -f  6)"-1. 

EXERCISES 

Differentiate  the  following  functions : 

1.   y=-^—-  Ans.  Dxy=   1  +  a?2-. 

1  t3 

2-   IT-TTtr  4-   2/  = 


l  +  «2  1  — a? 

„           1-1  x2 

3.    s  =  - 5.    y  = 


1  +  t  "      (1  -  a)2 

6.  ?/  =  aj(a  +  6a?)\     ^4rcs.     Dxy  =  [a  +  (n  -f  1) to] (a  +  foe)*-1. 

7.  y  =  — ,  where  m  is  a  positive  integer. 

xm 

l-\-x-\-x2  n  S  —  x^  +  Sx* 

8.  y==     t    -r —  9>  x 

a;  ar 

10.   Show  that  Formula  (12)  holds   when  n  is  a  negative 
integer. 


15.  *±£?. 

2  +  a 

16.  (a  +  bx  4-  ca2)' 
1 


Dif 

ferentiate  further 

11. 

(*+3)(2-*). 

12. 

(a  —  x)s. 

13. 

l-3x-x* 

^           9 

14. 

s 

17 


(1-ary 


18.    ,-?t=l 


(2s  +  3)! 
19.   Find  the  slope  of  the  curve  y  =  -  at  the  point  (1,  1). 


DIFFERENTIATION  OF   ALGEBRAIC   FUNCTIONS      25 

20.   Find  the  slope  of  the  curve 

240?/  =  (1  -  x)  (2  -  x)  (3  -  x)  (4  -x) 
at  the  point  x  =  0,  y  =  fa.  Ans.  — -  -^ . 

7.   Differentiation  of  Radicals.     Let  us  differentiate 
y  =  Va;. 


Here,  y0  =  V#0,  y0  +  Ay  =  V#0  4-  Aa;, 


Ay  __  V^o  +  Ax  —  V  %n . 

Ax  Ax 

We  cannot  as  yet  see  what  limit  the  right-hand  side  approaches 
when  Ax  approaches  0,  for  both  numerator  and  denominator 

approach  0,  and  -  has  no  meaning,  cf.  §  5.     We  can,  however, 

transform  the  fraction  by  multiplying  numerator  and  denomi- 
nator by  the  sum  of  the  radicals  and  recalling  the  formula  of 
Elementary  Algebra : 

a2  -  b2  =  (a  -  b)  (a  +  b). 

„         A?/      V#0  +  Ax  — '  V#„     V#0  +  Ax  -}-  V#0 

Thus  t~ r *  -— === — 

Aa  Ax  Va0  +  Ax  +  V#0 

_.  1    ,     (x0  +  Ax)—x0 1 

A#     ^/Xq  _|_  Ax  +  V^0      V#0  4-  A#  4-  W 

and  hence   lim  — ^  =  lim  —  ■ —  = —  • 

a*=o  Ao;      ax= o  -yJXo  -|-  Aa?  4-  Va?0      2  V»0 

Dropping  the  subscript,  we  have  : 


(16)  Dx^x  = 


1 


2Vx 
We  can  now  differentiate  a  variety  of  functions. 


26  CALCULUS 


Example.     To  differentiate 


y  =  Va2  —  x2. 

Introduce  a  new  variable 

z^a2-^ 

(I      / 
and  then  apply  Theorem  Y :  * 

2/  =  Va, 

A2/  =  AV^=AV^.Z>^=-^r-(-2a;) 


2Vz  Va2— a2 


Hence  Z>xVa2-a2  =        ~" g 

Va2  —  x2 


EXERCISES 

Differentiate  the  following  functions : 

1.  y  =  Va2  +  x2.  6.    y  =  —->     Ans. j 

Vx  2x? 

2.  u  =  Va  +  x  4-  Va  —  #.  7     —  . 


Va2  —  a?2 


3.    y  =  a?Vl— a?.  8.   y  =  ^/'2mx. 

x-\-\ 

v  a? 


5.       ,  Ans.    —  • 

vr^?         (Vi-z2)3       '  vi+x 

*  The  student  should  observe  that  Theorem  V  is  not  dependent  on  the 
special  letters  used  to  designate  the  variables.    Thus  if,  as  here, 

y=f(z)  and  z  =  \J/ (x) 

we  have  Dxy  =  Dzy  •  Dzz. 


DIFFERENTIATION  OF   ALGEBRAIC   FUNCTIONS      27 
8.    Continuation :   xn,  n  Fractional. 

•  The  Laivs  of  Fractional  Exponents.  Let  n=p/q,  where  p 
and  q  are  positive  integers  prime  to  each  other,  and  consider 
the  function 

p 
(17)  y  =  xn  =  xq 

for  positive  values  of  x.  If  q  is  odd,  the  function  is  single- 
valued  ;  but  if  q  is  even,  there  are  two  ^th  roots  of  x,  and  we 
might  define  the  function  of  (17)  to  be  double-valued,  namely, 
as  ±(-\/x)p.  This  is.  however,  inexpedient,  and  usage  has 
determined  that  the  notation  (17)  shall  be  defined  to  mean  the 
positive  root :  p 

x*  =  (yx)p. 

If  n  is  a  negative  fraction,  n  =  —  m,  then 

xn  =  —        Moreover,     a0  =  1. 

xm 

In  Elementary  Algebra  the  following  laws  of  exponents  are 
established : 


(A) 


These  laws  hold  without  exception  when  a  and  b  are  both 
positive  and  m  and  n  are  any  positive  or  negative  integers  or 
fractions,  inclusive  of  0. 

Graph  of  the  Function  xn.  When  n  is  an  integer,  n  =  1,  2, 
3,  •••,  10,  •••,  the  graphs  are  as  indicated  in  Fig.  3;  for  the 
slope  of  the  curve  (17) : 

tanr  —  Dxy  —  nxn~Y 

is  positive  and  increases  steadily  as  x  increases  if  n  >  1. 


I. 

am  -an  =  am+n ; 

II. 

(am)n  =  amn ; 

III. 

an  •  bn  =  (ab)n. 

28  CALCULUS 

Consider  next  the  case  that  p  =  1,  q  >  1.     Here 


(18) 


y  =  x< 


or 


<B  =  y 


and   so,  when   q  =  2,  3,  •  •,  10,  •  ••,  we  get  the  same  graphs  as 
I 


y=l 


=  10 
n- 


nf\ 


.^i 


x  =  l 


Fig.  3 


when  n  is  an  integer,  only  drawn  with  y  as  the  axis  of  abscissas 
and  x  as  the  ordinate,  cf.  Fig.  3.     Thus  any  one  of  the  latter 


DIFFERENTIATION  OF   ALGEBRAIC   FUNCTIONS      29 

curves,  as  y  =  a?1/3,  is  obtained  from  the  corresponding  one  of 
the  former  curves,  y  =  x3,  by  reflecting  this  curve  in  the  bisec- 
tor of  the  angle  made  by  the  positive  coordinate  axes. 

The  general  case,  n  =  p/q,  will  be  taken  up  at  the  close  of 
the  paragraph. 

Differentiation  ofxn.  Let  us  first  find  the  slope  of  the  curve 
(18).  If  <r  denotes  the  angle  between  its  tangent  and  the 
axis  of  y,  then 

tan  a  =  Dyx  =  qy*~\ 

Now  <r  is  the  complement  of  t,  and  so* 

1 


tanr  = 


tan<r 
Hence  Dxy  = — —  =  -yl~q. 

qy"'1    q 

*  This  is  equivalent  to  the  relation  : 

Dxy  =  -L- 
Dyx 

It  is  easy  to  give  a  proof  of  this  relation  as  follows.     If 

y  =/(*) 

is  any  continuous  function  of  x  whose  inverse  function : 

x  =  t(y) 

is  single-valued  near  the  point  (zo,  yd)  at  which  we  are  considering  the 
derivatives,  then 

Ax     Ax 
Ay 
and  hence,  if  lira  Ax/ Ay  9^0, 

lira  *V  = I or  j)xy=J_  q#e.d. 

***>Ax       Hm  Ax  y      Dvx" 


30  CALCULUS 

Keplacing  y  by  its  value,  a^/?,  we  have  : 

i  l  l-9  1 

yl~«=:(xq)    9  =  x~=oc?    , 


and  thus 


1      1   §-' 


(19)  Dxxq=-xq    • 

Q 

This  shows  that  Formula  (12)  holds  even  when  n  =  1/q. 

Turning  now  to  the  general  case : 

p 
y  =  x«, 
i 
let  z  =  xq ;  ?/  =  zp. 

Then  by  Theorem  V,  §  5,  and  (19)  : 

1    ~i 

Dxy  =  Dxzp  =  Dzzp  >Dxz  =  pzp~x  •  -xq   , 

zp-i=(x^y 


Hence  Dxy=*xq  -  xq    =£-xq   , 

or 

(20)  ZJ.afssnaj-1, 

when  n  is  any  positive  integer  or  fraction. 

If  n  is  a  negative  integer  or  fraction :  n  =  —  m, 

then  ajw  =  — , 

and  hence  £cw  can  be  differentiated  by  the  aid  of  Theorem  IV, 

a;m  xrm 

or  -Dxa?n  =  wajn-1. 

Consequently  Formula  (12)  holds  for  all  commensurable  values 
of  n.  We  shall  show  later  that  it  holds  for  incommensurable 
values,  too,  and  thus  is  true  for  any  fixed  value  of  n. 


DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS       31 

i 

If   n  =  i,  we  have 


2> 

X 


2Vx  ' 


which  agrees  with  (16). 

.  Example.     To  differentiate 

^  =  Va2  —  a^. 
Let  z  =  a2  —  x2. 

Then  2/  =  z% 

Dxy  =  Dxz?  =  Dzz^Dxz  =  \z-*(-2x)  = 


3(Va2-^)2 
Inequalities  for  xn.    If  n  is  a  positive  integer,  then 

r    xn  >  1  when  #  >  1 ; 

1    xn<l  when         0<a<l. 

The  same  relations  hold  when  n  =p/q  is  a  positive  fraction. 

For,  suppose 

i_ 

y  =  xq<  1  when  x  >  1. 

Then,  by  I,  2/*  <  1- 

But  yq  —  x,  and  #<1   is   contrary  to   hypothesis.     Similarly 
when  x  <1. 

Finally,  relations  I.  hold  when  n=p/q  is  any  positive  frac- 
tion.    For 

xq  =  (iC9)  , 
and  if  a?>l,  then 

1  /  \p 

xq>l  and  hence  (x9)  >1. 

Similarly  for  a?  <  1. 

When,  however,  n  <  0,  the  first  inequality  sign  in  each  line 
of  I.  must  be  reversed  and  the  value  x  =  0  excluded.  The 
function  xn  is  nevertheless  always  positive  when  x  >  0. 


32  CALCULUS 

Theorem.    If  n'  >  n,  then 

(a)  xn'  >  xn        when        x  >  1 ; 

(b)  xn' <xn        when        x<l. 
Let  n'  =  n  +  h.     Then 

a?n'  -  xn  =  #n+A  —  ccn  =  af(a*  —  1). 

Since  /i>0,  we  see  from  the  relations  I.  that  when  a?>l,  this 
last  expression  is  positive  ;  when  x  <1,  it  is  negative.  Hence 
the  theorem. 

Graph  of  the  Function  xn ;  Conclusion.  From  the  theorem 
just  established  it  follows  that  the  graphs  of  xn  for  different 
positive  fractional  values  of  n  lie  as  suggested  in  Fig.  3, 
namely :  they  all  pass  through  the  origin  and  the  point  (1,  1), 
and  they  have  no  other  point  of  intersection.  Their  slope  is 
always  positive.  Of  two  graphs  corresponding  to  n  and  n'  >  n, 
the  latter  lies  below  the  former  when  x  <  1 ;  above  it  when 
a?>l* 

The  student  is  requested  to  write  out  similar  statements  for 
the  case  that  n  <  0,  and  to  draw  the  graphs.  It  is  better,  how- 
ever, not  to  complicate  Fig.  3  with  these  latter  graphs,  but  to 
deduce  them  when  needed  from  Fig.  3.  Thus  confusion  will 
be  avoided.  Fig.  3  should  be  permanently  visualized.  The 
student  should  construct  such  a  figure  for  himself  accurately 
on  coordinate  paper,  using  the  tables  of  squares  and  cubes. 

*  These  curves  penetrate  every  part  (a)  of  the  square,  the  coordinates 
of  whose  points  (x,  y)  satisfy  the  relations  : 

0<s<l,  0<y<l, 

and  (6)  of  the  interior  of  the  right  angle  of  Fig.  3  : 

l<x,  1<». 

When  we  include  the  curves  for  which  n  is  positive  and  incommensurable, 
the  complete  family  y  =  xn  thus  obtained  just  fills  these  regions  without 
overlapping.    For  a  proof  of  these  statements  see  the  Appendix. 


DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS      33 

EXERCISES 

Differentiate  the  following  functions : 

1.  y  =  10x%  —  4af*  —  1.  4.  y  =  ^ax2. 

2.  y=zx*-x-*  +  w.  5     v  =  l-x~1/2 

'   y  x^ 

Vic  6.   y  =  x^/2x. 

1  r,  2x 

</a*-x*  3(a2-a^ 

8.    y  =  x(l-x*)*\  9.   4^+JL--- 

io.  (a^  +  ijV^y.   ^tns.  7*/4-V-i. 

2(f-y)? 

11     JpL=*L.  13     ^H-**, 

'    \(l  +  z)2  13'    ^^ 

12.    aa  —  ax  +  a.  14.    aa+6  —  af"6. 


15.  <*-*)'■    ^  SoV^-tf 


16.       ;a~a?-.  17.    r  =  Va$. 


V2 


ax  —  a? 


18     Vtt  —  #  4-  Va  +  a?  ^         a2  +  aVa2  —  ar* 

V«  —  x  —  Va  +  aj  x2s/a2  —  a;2 

19.  Find  the  slope  of  the  curve  y  =  a?*  in  the  point  whose 
abscissa  is  2,  correct  to  three  significant  figures. 

Ans.   tan  r  =  .115. 

20.  If  jw"=xC;  find  Al>- 

s-1 


21.    If  y^x  =  l  +  x,  ftnd  Dxy.  -4w*. 


2a;VaJ 


34  CALCULUS 

9.   Differentiation  of  Algebraic  Functions.     When  x  and  y 
are  connected  by  such  a  relation  as 

x2  -f-  y2  =  a2 

or  Xs  —  xy  -f  y5  =  0 

or  xysmy  =  x  +  ylogx, 

i.e.  if  y  is  given  as  a  function  of  x  by  the  equation 

or  its  equivalent,  ®(x,  y)  =  ty(x,  y),  where  neither  <l>  nor  ^  re- 
duces to  y,  then  y  is  said  to  be  an  implicit  function  of  x.  If  we 
solve  the  equation  for  y,  thus  obtaining : 

y  =/(«), 

2/  thereby  becomes  an  explicit  function  of  a?.  It  is  often  difficult 
or  impossible  to  effect  the  solution ;  but  even  when  it  is  possible, 
it  is  usually  easier  to  differentiate  the  function  in  the  implicit 
form.  Thus  in  the  case  of  the  first  example  we  have,  on  dif- 
ferentiating the  equation  as  it  stands  with  respect  to  x: 

Dxx2  +  Dxy2  =  Dxa2 

or  2x  +  2yDxy  =  0. 


Hence  Dxy  =  —  -  • 

y 

If  we  differentiate  the  second  equation  in  a  similar  manner, 
we  get : 

3  x2  —  x  Dx  y  —  y  +  5  yK  Dx  y  =  0. 

Solving  for  Dxy,  we  obtain : 

r,  3  x2  —  y 

by 4  —  x 


DIFFERENTIATION  OF  ALGEBRAIC   FUNCTIONS       35 

When  F(x,  y)  is  a  polynomial  in  x  and  y,  the  function  y, 
denned  by  the  equation 

F(x,y)  =  0 

is  called  an  algebraic  function.  Thus  all  polynomials  and 
fractional  rational  functions  are  algebraic.  Moreover,  all 
functions  expressed  by  radicals,  as 


2/=Va2-x2  or  y  =  ^ljz£?_^4-V*, 

*  1  -f-X 

are  algebraic,  for  the  radicals  can  be  eliminated  and  the  result- 
ing equation  brought  into  the  above  form.  But  the  converse 
is  not  true:  not  all  algebraic  equations  can  be  solved  by  means 
of  radicals. 

It  can  be  shown  that  an  algebraic  function  in  general  is 
continuous.  In  case  the  function  is  multiple-valued  it  can  be 
considered  as  made  up  of  a  number  of  branches,  each  of  which 
is  single-valued  and  continuous.  Assuming  this  theorem  we 
can  find  the  derivative  of  an  algebraic  function  in  the  manner 
illustrated  in  the  foregoing  differentiations. 

On  the  assumption  just  mentioned  a  short  proof  of  Formula 
(12),  §  2,  can  be  given  for  the  case  that  n—p/q.     Since 

p 
y  =  xq,  we  have:         y9  =  xp. 

Differentiate  each  side. with  respect  to  x: 

Dxyq  =  Dxxp  =  pxr-\ 

Now  Dmtf  =  Dyy*  •  Dxy  =  <nTlD,y. 

Hence  Dmy  -«^-«  J^l .««?-. 

qyq~l    q  |<«-i>    q 

X? 

The  proof  of  this  formula  which  we  gave  in  §  8  does  not 
depend  on  the  above  assumption,  but  is  a  complete  proof. 


36  CALCULUS 

EXERCISES 

1.  Differentiate  y  in  both,  ways,  where 

xy  +  4:y  =  3x, 
and  show  that  the  results  agree. 

2.  The  same  for  y2  —  2mx. 

3.  Find  the  slope  of  the  curve 

x4-2xy2  +  y5=13 
at  the  point  (2,  1).  Ans.   10. 

4.  Show  that  the  curves 

Sy  =  2x  +  x4fy  2y  +  3x  +  y5  =  x3y 

intersect  at  right  angles  at  the  origin. 


CHAPTER   III 

APPLICATIONS 

1.   Tangents  and  Normals.     The  equation  of  a  line  passing 
through  a  point  (#0,  y0)  and  having  the  slope  a  is 

y-yo  =  Hx~xo), 

and  the  equation  'of  its  perpendicular  through  the  same  point  is 
y-yo=-~(x-x0)  or  x  —  x0  +  \(y  —  y0)  =  0. 

A 

Since  the  slope  of  a  curve 

y=f(x)  or  F(x,y)=0. 

in  the  point  (x0,  y0)  is  [Dxy~\x=x  ,  the  equation  of  the  tangent 
line  in  that  point  is 

(1)  2/  — 2/o=  [Dxy~]x=:XQ(x -  x0). 
Similarly,  the  equation  of  the  normal  is  seen  to  be : 

(2)  2/-2/o=-jt— —  (x-xo)    or  x-  x0  +  [2>»y],.s(y-3fii)-»0. 

Example   1.     To  find  the  equation  of  the  tangent  to  the 
curve 

y  =  x? 

in  the  point  x  =  J,  y  =  J.     Here 

Dx2/  =  3^         [Z>x2/]x=x=[3^]x  =  i  =  |. 

Hence  the  equation  of  the  tangent  is 

2/  —  |  =  f(cc  —  i)  or  3a;— 4?/  — 1  =  0. 

37 


38  CALCULUS 

Example  2.     Let  the  curve  be  an  ellipse : 

a2^b2 

Differentiating  the  equation  as  it  stands,  we  get : 

2x  .  2y  n         a  ^  b2x 

—  +  -rf  Dxy  =  0,  J),?/  -  ~  -T-  * 

a2       b1  a-y 

Hence  the  equation  of  the  tangent  is 

0  Xt\  /  \ 

«l/o 
This  can  be  transformed  as  follows  : 

a2y0y  —  a2y02  =  —  b2x0x  +  b2x02, 
b2x0x  +  a2y0y  =  a2y02  +  b2x<?  =  a2b2, 

®o%  ■  yo.V==i 
"a2       62 

EXERCISES 

1.  Find  the  equation  of  the  tangent  of  the  curve 

y  =  x3  —  x 
at  the  origin ;  at  the  point  where  it  crosses  the  positive  axis 
of  x.  Ans.    x  +  y  —  0;     2x  —  y  —  2  =  0. 

2.  Find  the  equation  of  the  tangent  and  the  normal  of  the 
circle 

x2  -f-  y2  =  4 

at  the  point  (1,  V3)  and  check  your  answer. 

3.  Show  that  the  equation  of  the  tangent  to  the  hyperbola 

a2     b2 
at  the  point  (x0,  y0)  is 

a2        b2 

4.  Find  the  equation  of  the  tangent  to  the  curve 

x3  -f  y3  =  a2  (x  —  y) 
at  the  origin.  Ans.   x  =  y. 


APPLICATIONS  39 

5.  Show  that  the  area  of  the  triangle  formed  by  the  coordi- 
nate axes  and  the  tangent  of  the  hyperbola 

xy  =  a2 
at  any  point  is  constant. 

6.  Find  the  equation  of  the  tangent  and  the  normal  of  the 

curve 

x6  =  a?y2 

in  the  point  distinct  from  the  origin  in  which  it  is  cut  by  the 
bisector  of  the  positive  coordinate  axes. 

7.  Show  that  the  portion  of  the  tangent  of  the  curve 

at  any  point,  intercepted  between  the  coordinate  axes  is 
constant. 

8.  The  parabola  y2  =  2ax  cuts  the  curve 

x3  —  3aa?2/  +  2/3  =  0 

at  the  origin  and  at  one  other  point.  Write  down  the  equa- 
tion of  the  tangent  of  each  curve  in  the  latter  point. 

9.  Show  that  the  curves  of  the  preceding  question  intersect 
in  the  second  point  at  an  angle  of  32°  12'. 

2.  Maxima  and  Minima. 

Problem.  Find  the  most  advantageous  length  for  a  lever, 
by  means  of  which  to  raise  a  weight  of  100  lbs.  (see  Fig.  4),  if 
the  distance  of  the  weight  from  <0| 

the  fulcrum   is   2   ft.    and   the   0^  *        P]\ 

lever  weighs  3  lbs.  to  the  foot.       ~   ^w)  [  Q 

It  is  clear  that,  if  we  make  6x 

the  lever  very  long,  the  increased 

weight  of  the  lever  will  more  than  compensate  for  the  gain  in 
the  leverage,  and  so  the  force  P  required  to  raise  the  weight 
will  be  large.  On  the  other  hand,  if  we  make  the  lever  very 
short,  say  3  ft.  long,  the  force  required  to  lift  the  lever  is 
slight,  but  there  is  little  advantage  from  the  leverage ;  and  so, 


40 


CALCULUS 


again,  the  force  P  will  be  large.  Evidently,  then,  there  is  an 
intermediate  length  that  will  give  the  best  result,  i.e.  for 
which  P  will  be  least. 

Let  x  denote  the  half-length  of  the  lever.  Then  P  is  a  func- 
tion of  x.  If  we  determine  this  function,  i.e.  express  P  as  a 
function  of  x,  we  can  plot  the  graph  and  see  where  it  comes 
nearest  to  the  axis  of  x.  Now  the  moments  of  the  forces  that 
tend  to  turn  the  lever  about  the  fulcrum  O  in  the  clockwise 
direction  are : 

(a)     100  x  2,  due  to  the  weight  of  100  lbs. ; 
(6)     3  x  2x  x  x,  due  to  the  weight  of  the  lever. 

Hence  their  sum  must  equal  the  moment  of  P  in  the  opposite 
direction,  namely,  Px  2x,  and  thus 


200  +  6x2  =  2Px, 


P  = 


100  4-  3x2 


Let  us  try  a  few 
values  of  x  and  see 
what  the  correspond- 
ing values  of  P  are  : 


X 

P 

3 

421 

4 

37 

5 

35 

6 

34f 

7 

35f 

8 

36$ 

9 

38J 

Fig.  5 


From  these  figures  it  appears   that  the  best  length   corre- 
sponds to  a  value  of  x  between  5  and  7.     By  computing  P  for 


APPLICATIONS  41 

a  sufficiently  large  number  of  values  of  x  intermediate  between 
these  two  numbers,  we  could  evidently  approximate  to  the  best 
value  as  closely  as  we  wished.  Can  we  not,  however,  with  the 
aid  of  the  Calculus,  save  ourselves  the  labor  of  these  computa- 
tions ?  Looking  at  the  graph  of  the  function,  we  see  that  the 
value  of  x  we  want  to  find  is  that  one  which  corresponds  to  the 
smallest  ordinate.  This  point  of  the  curve  is  characterized  by 
the  fact  that  the  tangent  is  here  parallel  to  the  axis  of  x  and 
hence  the  slope  of  the  curve  is  0  : 

tanT  =  Z>xP=0. 

Let  us  compute,  then,  DXP  and  set  it  equal  to  0: 

DXP=  -^  +  3  =  0, 

^100  L^=5.77. 

3  '  ys 

Consequently  the  best  length  for  the  lever  is  2x  =  11.4  ft.,  the 
corresponding  value  of  P  being  34.6  lbs. 

Example.  A  box  is  to.be  made  out  of  a  square  piece  of  card- 
board 4  in.  on  a  side,  by  cutting  out  equal  squares  from  the 
corners  and  turning  up  the  sides.  Find  the  dimensions  of  the 
largest  box  that  can  be  made  in  this  way. 

First  express  the  volume  V  oi  the  box  in  terms  of  the  length 
x  of  a  side  of  one  of  the  squares  cut  out,  and  plot  the  graph  of 
V,  thus  determining  approximately  the  best  value  for  x.  Then 
solve  by  the  Calculus. 

The  foregoing  examples  suggest  a  simple  test  for  a  maximum 
or  a  minimum. 

Test  for  a  Maximum.     If  the  f miction 

is  continuous  ivithin  the  interval  a<.x<b  and  has  a  larger  value 
at  one  of  the  intermediate  points  than  it  does  at  or  near  each  of 
the  ends,  then  it  has  a  maximum  at  some  point  x  =  x0  within  the 
interval. 


42  CALCULUS 

Under  the  above  conditions  we  shall  have : 

Dxy  =  0      when      x  =  x0. 

And  if  this  equation  has  only  one  root  in  the  above  interval,  then 
this  root  must  be:  x  =  x0* 

The  test  for  a  minimum  is  similar,  the  words  "larger" 
and  "maximum"  being  merely  replaced  by  "smaller"  and 
"  minimum." 

EXERCISES 
li.   Find  the  least  value  of  the  function 

y  =  x2  -f  6x  -f- 10.  Ans.   1. 

2.  What  is  the  greatest  value  of  the  function 

y  —  Sx  —  x* 
for  positive  values  of  x  ? 

3.  For  what  value  of  x  does  the  function 

12Va 

l  +  4a; 

attain  its  largest  value  ?  Ans.    x  =  \. 

*  It  is  true  that  there  are  exceptions  to  the  test  as  stated,  for  the  graph 
of  a  continuous  function  may  have  sharp  corners,  as  shown  in  Fig.  A. 
At  such  a  point,  however,  the  function  has  no 
derivative,  since  Ay/ Ax  approaches  no  limit, 
as  at  x  =  xi,  or  it  approaches  one  limit  when 
Ax  approaches  0,  passing  only  through  positive 
values,  and  another  limit  when  Ax  approaches 


x, 


% —  "^   /. —    0  from  the  negative  side,  as  at  x  =  x2.    If, 


pIQ   a  then,  we  add  the  further  condition  to  our  test, 

that  f{x)   shall   have    a  derivative  at  each 
point  of  the  interval,  no  exception  can  possibly  occur. 

This  condition  is  obviously  fulfilled  when,  as  is  usually  the  case  in  prac- 
tice, we  are  able  to  compute  the  derivative, 


APPLICATIONS 


43 


4.   At  what  point  of  the  interval  a<  x  <b,  a  being  positive, 
does  the  function 

x 


attain  its  least  value  ? 


(x  —  a)  (6  —  x) 


Ans.  x  =  Va6. 


5.  The  legs  of  an  isosceles  triangle  are  each  6  in.  long. 
How  long  must  the  base  be  made  in  order  that  the  area  may 
be  a  maximum  ?  Ans.    6  V2  =  8 \  in. 

6.  A  two-acre  pasture  in  the  form  of  a  rectangle  is  to  be 
fenced  off  along  the  bank  of  a  straight  river,  no  fence  being 
needed  on  the  river  side.  What  must  be  its  shape  in  order 
that  the  fence  may  cost  as  little  as  possible  ? 

Ans.    It  must  be  twice  as  long  as  it  is  broad. 

3.  Continuation;  Auxiliary  Variables.  It  frequently  hap- 
pens that  in  formulating  a  problem  in  maxima  and  minima  it 
is  advisable  to  express  the  function  which  is  to  be  made  a 
maximum  or  a  minimum  in  terms  of  two  variables,  between 
which  a  relation  exists.  The  following  example  will  illustrate 
the  method. 

To  find  the  largest  rectangle  that  can  be  inscribed  in  a 
circle. 

It  is  evident  that  the  area  of  the  rectangle 
will  be  small  when  its  altitude  is  small  and  also 
when  its  base  is  short.  Hence  the  area  will  be 
largest  for  some  intermediate  shape. 

From  the  equations  :    . 

u  =  kxy,  x2-\-y2  =  a2, 

we  could  eliminate  y  and  thus  express  u  as  a  function  of  x. 
In  practice,  however,  it  is  usually  better  not  to  eliminate,  but 
to  differentiate  the  equations  as  they  stand : 

D.u  =  4:(y  +  xDxy)  =  0,  2x  +  2yDxy  =  0. 


x   \ 

1 

?"'"V\ 

1 

1 

Fig.  G 


Hence 


D*y 


44  CALCULUS 

Substituting  this  value  in  the  first  equation,  we  get 


x- 


y =  0        or        y  =  x, 

y 

i.e.,  the  maximum  rectangle  is  a  square. 


EXERCISES 

1.'  A  100-gallon  tank  is  to  be  built  with  a  square  base  and 
vertical  sides,  and  is  to  be  lined  with  copper.  Find  the  most 
economical  proportions. 

Ans.    The  length  and  breadth  must  each  be  double  the  height. 

2.  Find  the  greatest  cylinder  of  revolution  that  can  be 
inscribed  in  a  given  cone  of  revolution. 

3.  What  is  the  cylinder  of  greatest  convex  surface  that  can 
be  inscribed  in  the  same  cone  ? 

A  4.  Find  the  volume  of  the  greatest  cone  of  revolution  that 
can  be  inscribed  in  a  given  sphere. 

5.  Find  the  most  economical  proportions  for  a  cylindrical 
tin  dipper  which  is  to  hold  a  pint.  Ans.  h  =  r. 

*  6.  What  ought  to  be  the  shape  of  a  tomato  can  to  hold  a 
quart  and  to  require  as  little  tin  as  possible  for  its  manu- 
facture ? 

7.  If  the  top  and  bottom  of  the  can  are  cut  from  sheets  of 
tin  in  such  a  way  that  a  regular  hexagon  is  used  up  each  time 
and  the  waste  is  a  total  loss,  what  will  then  be  the  best 
proportions  ? 

■*  8.  A  Norman  window  consists  of  a  rectangle  surmounted  by 
a  semicircle.  If  the  perimeter  of  the  window  is  given,  what 
must  be  its  proportions  in  order  to  admit  as  much  light  as 
possible  ?  Ans.   Breadth  and  height  equal. 

9.  Work  the  last  two  questions  of  the  preceding  Exercises 
by  the  present  method. 


APPLICATIONS  45 

^  10.' Assuming  that  the  stiffness  of  a  beam  is  proportional 
to  its  breadth  and  to  the  cube  of  its  depth,  find  the  dimen- 
sions of  the  stiffest  beam  that  can  be  sawed  from  a  log  one  foot 
in  diameter. 

./ 

11.  If  the  cost  per  hour  of  running  a  certain  steamboat  in 

still  water  is  proportional  to  the  cube  of  the  velocity,  find  the 
most  economical  rate  at  which  to  run  the  steamer  up  stream 
against  a  four-mile  current.  Ans.  6  m.  per  h. 

12.  The  gate  in  front  of  a  man's  house  is  20  yds.  from  the 
car  track.  If  the  man  walks  at  the  rate  of  4  miles  an  hour  and 
the  car  on  which  he  is  coming  home  is  running  at  the  rate  of 
12  miles  an  hour,  where  ought  he  to  get  off  in  order  to  reach 
home  as  early  as  possible  ? 

13.  If  the  cost  per  hour  for  the  fuel  required  to  run  a  given 
steamer  is  proportional  to  the  cube  of  her  speed  and  is  $  20  an 
hour  for  a  speed  of  10  knots,  and  if  other  expenses  amount  to 
$135  an  hour,  find  the  most  economical  rate  at  which  to  run 
her.  Ans.   15  knots  an  hour. 

v  14.  A  telegraph  pole  at  a  bend  in  the  road  is  to  be  sup- 
ported from  tipping  over  by  a  stay  20  ft.  long  fastened  to  the 
pole  and  to  a  stake  in  the  ground.  How  far  from  the  pole 
ought  the  stake  to  be  driven  in  ? 

15.  How  much  water  should  be  poured  into  a  cylindrical 
tin  dipper  in  order  to  bring  the  centre  of  gravity  as  low  down 
as  possible  ? 

\/  16.  *  A  man  is  in  a  row  boat  3  miles  from  the  nearest  point 
A  of  a  straight  beach.  He  wishes  to  reach  a  point  of  the 
beach  5  miles  from  A  in  the  shortest  possible  time.  If  he 
can  walk  at  the  rate  of  4  miles  an  hour,  but  can  row  only 
3  miles  an  hour,  what  point  of  the  beach  ought  he  to  row 
for? 


46  CALCULUS 

4.  Velocity.  By  the  average  velocity  with  which  a  point 
moves  for  a  given  length  of  time  t  is  meant  the  distance  s 
traversed  divided  by  the  time : 

average  velocity  =  — 

Thus  a  railroad  train  which  covers  the  distance  between  two 
stations  15  miles -apart  in  half  an  hour  has  an  average  speed 
of  15/-J-  =  30  miles  an  hour. 

When,  however,  the  point  in  question  is  moving  sometimes 
fast  and  sometimes  slowly,  we  can  describe  its  speed  approxi- 
mately at  any  given  instant  by  considering  a  short  interval 
of  time  immediately  succeeding  the  instant  t0  in  question,  and 
taking  the  average  velocity  for  this  short  interval. 

For  example,  a  stone  dropped  from  rest  falls  according  to 

the  law : 

s  =  16Z2. 

To  find  how  fast  it  is  going  after  the  lapse  of  t0  seconds.     Here 

(1)  s0=16t02. 

A  little  later,  at  the  end  of  t'  seconds  from  the  beginning  of 
the  fall, 

(2)  s'  =  16t'2 

and  the  average  velocity  for  the  interval  of  t1  —  t0  seconds  is 

(3)  ^7=^  ft.  per  sec. 
W  t'-t0        * 

Let  us  consider  this  average  velocity,  in  particular,  after  the 
lapse  of  1  second : 

t0  as  1,  S0  =  16. 

Let  the  interval  of  time,  t'  —t0,  be  y1^  sec.     Then 
*'  =  16  x  l.l2  =  19.36, 
9o  =  3136==33_6ft  agea 


t'  - 10       .1 

Next,  let  the  interval  of  time  be  T^  sec.     Then  a  similar 
computation  gives,  to  three  significant  figures : 


APPLICATIONS  47 

j, — 7°  =  32.2  ft.  a  sec. 

And  when  the  interval  is  taken  as  y^g-  sec.,  the  average 
velocity  is  32.0  ft.  a  sec. 

Thus  we  can  get  at  the  speed  of  the  stone  at  any  desired 
instant  to  any  desired  degree  of  accuracy  by  direct  computa- 
tion; we  need  only  to  reckon  out  the  average  velocity  for  a 
sufficiently  short  interval  of  time  succeeding  the  instant  in 
question. 

We  can  proceed  in  a  similar  maimer  when  a  point  moves 
according  to  any  given  law.  Can  we  not,  however,  by  the  aid 
of  the  Calculus  avoid  the  labor  of  the  computations  and  at  the 
same  time  make  precise  exactly  what  is  meant  by  the  velocity 
of  the  point  at  a  given  instant  ?  If  we  regard  the  interval 
of  time  t'  — 10  as  an  increment  of  the  variable  t  and  write 
V  —  t0  =  At,  then  s'  —  s0  =  As  will  represent  the  corresponding 
increment  in  the  function,  and  thus  we  have : 

As 
average  velocity  =  — 

Now  allow  At  to  approach  0  as  its  limit.  Then  the  average 
velocity  will  in  general  approach  a  limit,  and  this  limit  we  take 
as  the  definition  of  the  velocity  v  at  the  instant  t0 : 

lim  (average  velocity  from  t  =  t0  to  t  =  t1) 

=  actual  velocity  at  instant  t  =  t0, 

or  v—  lim  —  =  D.s. 


4(i0Ai 


Hence  it  appears  that  the  velocity  of  a  point  is  the  time  deriva- 
tive of  the  space  it  has  travelled. 

Similarly,  the  rate  at  which  the  distance  between  two  points, 
one  or  both  of  which  are  moving,  is  changing  is  the  derivative 
of  their  distance  apart  with  respect  to  the  time;  see  Ex.  4 
below.  And  the  rate  at  which  any  quantity  is  changing  is  its 
time  derivative,  as  in  Ex.  3. 


48  CALCULUS 


EXERCISES 


1.  The  height  of  a  stone  thrown  vertically  upward  is  given 
by  the  formula:  s==48*-16*2. 

When  it  has  been  rising  for  one  second,  find  (a)  its  average 
velocity  for  the  next  T^-  sec. ;  (6)  for  the  next  y^-  sec. ;  (c)  its 
actual  velocity  at  the  end  of  the  first  second;  (d)  how  high  it 
will  rise. 

Ans.  (a)  14.4  ft.  a  sec. ;  (b)  15.84  ft.  a  sec. ;  (c)  16  ft.  a 
sec. ;  (d)  36  ft. 

2.  A  man  6  ft.  tall  walks  directly  away  from  a  lamp-post 
10  ft.  high  at  the  rate  of  4  miles  an  hour.  How  fast  is  the 
further  end  of  his  shadow  moving  along  the  pavement  ? 

Ans.   10  miles  an  hour. 

3.  Find  the  rate  at  which  the  shadow  in  the  preceding 
problem  is  lengthening. 

^•0-    The  rays  of  the  sun  make  an  angle  of  30°  with  the  hori- 

f'zon.     A  ball  is  thrown  vertically  upward  to  a  height  of  64  ft. 

How  fast  is  its  shadow  on  the  ground  travelling  just  before 

the  ball  strikes  the  ground  ?  Ans.   Ill  ft.  a  sec. 

"■"■— *Q.  Two  ships  start  from  the  same  port  at  the  same  time. 
One  ship  sails  east  at  the  rate  of  9  knots  an  hour,  the  other 
south  at  the  rate  of  1 2  knots.  H6w  fast  are  they  separating 
at  the  end  of  2  hours  ?  Ans.    15  knots  an  hour. 

6.  If  in  the  preceding  question  the  first  ship  starts  an  hour 
ahead  of  the  second  ship,  how  fast  will  they  be  separating  an 
hour  after  the  second  ship  leaves  port  ? 

7.  One  ship  is  20  miles  due  north  of  another  ship  at  noon, 
and  is  sailing  south  at  the  rate  of  10  knots  an  hour.  The  sec- 
ond ship  sails  west  at  the  rate  of  12  knots  an  hour.  How 
long  will  the  ships  continue  to  approach  each  other? 


APPLICATIONS 


49 


8.  A  stone  is  dropped  into  a  placid  pond  and  sends  out  a 
series  of  concentric  circular  ripples.  If  the  radius  of  the 
outer  ripple  increases  steadily  at  the  rate  of  6  ft.  a  sec,  how 
rapidly  is  the  area  of  the  water  disturbed  increasing  at  the 
end  of  2  sec.  ?  Ans.   452  sq.  ft.  a  sec. 

9.  A  man  is  walking  over  a  bridge  at  the  rate  of  4  miles  an 
hour,  and  a  boat  passes  under  the  bridge  immediately  below  him 
rowing  8  miles  an  hour.  The  bridge  is  20  ft.  above  the  boat. 
How  fast  are  the  boat  and  the  man  separating  3  minutes  later  ? 

5.  Increasing  and  Decreasing  Functions.  The  Calculus  af- 
fords a  simple  means  of  determining  whether  a  function  is 
increasing  or  decreasing  as  the  independent  variable  increases. 
Since  the  slope  of  the  graph  is  given  by  Dxy,  we  see  that, 
when  Dxy  is  positive,  y  increases  as  x  increases,  but  when  Dxy 
is  negative,  y  decreases  as  x  increases.  Fig.  7  shows  the  graph 
in  general  when  Dxy  is  positive. 

y 


Fig.  7 


Theorem  :    When  x  increases,  then 
(a)     if  Dxy>0,  y  increases ; 

Qj)     if  Dxy  <  0,  y  decreases. 

As  an  application  consider  the  condition  that  a  curve  y  =f(x) 
have  its  concave  side  turned  upward,  as  in  Fig.  8.  The  slope 
of  the  curve  is  a  function  of  a?:  y 

tanr  =  <f>(x). 

sider  the  tangent  line  at  a  vari- 
able point  P.     If  w    think  of  7'  as     ^ 
tracing  out  the  r  tA- ve  and  carrying  FlG<  8 


50 


CALCULUS 


the  tangent  along  with  it,  the  tangent  will  turn  in  the  counter 
clock-wise  sense,  the  slope  thus  increasing  algebraically  as  x 
increases,  whenever  the  curve  is  concave  upward.  And  con- 
versely, if  the  slope  increases  as  x  increases,  the  tangent  will 
turn  in  the  counter  clock-wise  sense  and  the  curve  will  be  con- 
cave upward.     Now  by  the  above  theorem,  when 

Dx  tan  t  >  0, 

tanr  increases  as  x  increases.  Hence  the  curve  is  concave 
upward,  when  Dx  tan  t  is  positive. 

The  derivative  Dx  tan  t  is  the  derivative  of  the  derivative 
of  y.     This  is  called  the  second  derivative  of  y  : 

(read :  "  D  x  second  of  y  ").# 

The^test  for  the  curve's  being  concave  downward  is  obtained 
in  a  similar  manner,  and  thus  we  are  led  to  the  following 
important  theorem. 

Test  for  a  Curve's  being  Concave  Upward,  etc.  The 
curve  y  =f(x) 

is  concave     upward  ivhen     Dx2y>0; 

concave  downward        when    Dx2y<0. 


Fig.  9 


*  The  derivative  of  the  second  derivative,  Dx  v  /*2  y) ,  is  called  the  third 
derivative  and  is  written  Dxzy,  and  so  on. 


APPLICATIONS  51 

A  point  at  which  the  curve  changes  from  being  concave  up- 
ward and  becomes  concave  downward  (or  vice  versa)  is  called 
a  point  of  inflection.  Since  D*y  changes  sign  at  such  a  point, 
this  function  will  necessarily,  if  continuous,  vanish  there. 
Hence : 

A  necessary  condition  for  a  point  of  inflection  is  that 

Dx2y  =  0. 

6.  Curve  Tracing.  In  the  early  work  of  plotting  curves 
from  their  equations  the  only  way  we  had  of  finding  out  what 
the  graph  of  a  function  looked  like  was  by  computing  a  large 
number  of  its  points.  We  are  now  in  possession  of  powerful 
methods  for  determining  the  character  of  the  graph  with 
scarcely  any  computation.  For,  first,  we  can  find  the  slope  of 
the  curve  at  any  point ;  and  secondly  we  can  determine  in 
what  intervals  it  is  concave  upward,  in  what  concave  down- 
ward. 

As  an  example  let  us  plot  the  graph  of  the  function  : 


(1) 

y  = 

-.  x3  +  px  +  q. 

Consider  first  the 

case  q  = 

=  0: 

(2) 

y  = 

:  Xs  -J-  pX. 

Here 

Dty  = 

>&*+i* 

DJy- 

=  6x. 

From  the  last  equation  it  follows  that  the  curve  is  concave 
upward  for  all  positive  values  of  x.  Moreover,  when  x  be- 
comes positively  infinite,  Dxy  also  becomes  positively  infinite, 
no  matter  what  value  p  may  have.  The  curve  goes  through 
the  origin  and  its  slope  there  is 


52 


CALCULUS 


Hence  the  graph  is,  in  character,  for  positive  values  of  x  as 
shown  in  Fig.  10. 

2»0      * 


Fig.  10 

To  obtain  the  graph  for  negative  values  of  x  we  need  only 
observe  that  the  curve  is  symmetric  with  respect  to  the  origin. 
For,  if  (x,  y)  be  any  point  of  the  curve,  then  x1  —  —  x, 
y'  =  —  y  is  also  a  point  of  the  curve.  When,  therefore,  we 
have  once  plotted  the  curve  for  positive  values  of  x,  we  need 
only  to  rotate  the  graph  through  180°  about  the  origin  in  order 
to  get  the  remainder  of  the  curve. 

Finally  we  can  get  the  graph  of  (1)  from  that  of  (2)  by 
merely  shifting  the  axis  of  x  to  a  parallel  axis.  The  formulas 
for  this  transformation  are  : 

x  =  x',  y  =  y'-q. 

Thus  (2)  becomes  : 

y'  —  q  ==  x's  +  px'. 

Transposing  q  and  dropping  the  accents,  we  get  equation  (1). 
This  curve  is  symmetric  with  respect  to  the  point  x  =  0,y  =  $. 


EXERCISES 

Use  coordinate  paper  in  working  these  exercises. 

1.    Show  that  the  curve 

?  =      3 
V      3  +  x2 

is  concave  downward  in  the  interval  —  1  <  x  <  1  and  concave 
upward  for  all  other  values  of  x.   Find  its  slope  in  its  points  of 


APPLICATIONS  X""53 

inflection  and  draw  the  tangent  line  in  each  of  these  points. 
Hence  plot  the  curve. 

2.  Plot  the  curve 

4y  =  x*  -  6x?  +  8, 
determining  ^^* 

(a)  its  intersections  with  the  coordinate  axes ; 

(b)  the  intervals  in  which  it  is  concave  upward  and  those  in 
which  it  is  concave  downward ; 

(c)  its  points  of  inflection ; 

(d)  the  points  where  its  tangent  is  parallel  to  the  axis  of  x. 
Plot  the  points  (a),  (c),  (d)  accurately,  using  a  scale  of  2  cm. 

as  the  unit,  and  draw  accurately  the  tangent  in  each  of  these 
points.     Hence  construct  the  curve. 

Plot  the  following  curves : 

3.  10y  =  x*-12x  +  9. 

Note  that  the  curve  cuts  the  axis  of  x  in  the  point  x  =  3. 

4.  y  =  a?  +  2x2-13x  +  10.  9.    y2  =  x2  +  x^. 

5.  y  =  x  —  x*.  10.   y2=x(x  —  l)(x  —  2). 

T  1 

6.  y=  -  ^11.   y  =  z s' 


•      1  -fiC2  1  —x 

1  „  T2 

8.    2/  =  -^--  13.   y2  =  ~ 


1-x  '       l+x2 

7.   Relative  Maxima  and  Minima.     Points  of  Inflection.    A 

function 

(i)  y=f{x) 

is  said  to  have  a  maximum  at  a  point  x  =  x0  if  its  value  at  .^0  is 
larger  than  at  any  other  point  in  the  neighborhood  of  x0.  But 
such  a  maximum  need  not  represent  the  largest  value  of  the 


54 


CALCULUS 


function  in  the  complete  interval  a<±x<b,  as  is  shown  by 
Fig.  11,  and  for  this  reason  it  is  called  a  relative  maximum,  in 

distinction  from  a 

V  i  • 

maximum      maxi- 

morum,  or  an  ab- 
solute maximum. 

A  similar  defini- 
tion holds  for  a 
minim  urn,  the  word 
"  larger  "  being 
5^-  merely  replaced  by 
"  smaller." 

It  is  obvious  that 
a  characteristic  feature  of  a  maximum  is  that  the  tangent  is 
parallel  to  the  axis  of  x,  the  curve  being  concave  downward. 
Similarly  for  a  minimum,  the  curve  here  being  concave  upward. 
Hence  the  following 

Test  for  a  Maximum  or  a  Minimum.     If 

(a)  WdflUk-^  [A'}],-t<0, 

the  function  has  a  maximum  for  x  =  x0 ;  if 

(b)  [A2/l«„  =  0,  [A2t/]I=lo>0, 

it  has  a  minimum. 

This  condition  is  sufficient,  but  not  necessary.     Thus  the 
function 


(2) 


y  =  a? 


evidently  has  a  minimum  at  the  point  x  =  0.  But  here  Dx2y 
is  not  positive,  but  =  0.  Still,  the  above  test  is  adequate  for 
the  great  majority  of  cases  that  arise  in  practice.  We  shall 
obtain  a  more  general  test  later. 


Example.     Let 


y  =  o?-3a?  +  l. 


APPLICATIONS  55 

Here  Dxy  =  6x* -  6x  =  6x(x> - 1)  (x2  + 1), 

and  hence         Dxy  =  0        for         a;  =  —  1,  0,  1. 

Furthermore,  Dx2y  =  30  a!4  -  6. 

Thus 

[A,2#]x=-i=   24  >0,         .-.   x  =  —  1  gives   a  minimum; 

[Dx2y~\x=0   =— 6<0,         .*.   a?  =     0       "      "   maximum; 

\_Dxy~]x=x   =    24  >0,         .-.    &  =      1        "      "   minimum. 

As  a  further  application  of  the  test  just  found  let  us  obtain 
a  sufficient  condition  for  a  point  of  inflection.  We  have  seen 
that  a  necessary  condition  is  that  Dxy  =  0;  but  this  is  not 
sufficient,  as  the  example  of  the  function  (2)  above  shows.  A 
geometric  feature  characteristic  of  a  point  of  inflection  is  that 
the  tangent  ceases  rotating  in  one  direction  and,  turning  back, 
begins  to  rotate  in  the  opposite  direction.  Hence  the  slope  of 
the  curve,  tanT,  has  either  a  maximum  or  a  minimum  at  a 
point  of  inflection. 

Conversely,  if  tan  t  has  a  maximum  or  a  minimum,  the 
curve  will  have  a  point  of  inflection.  For,  suppose  tan  r  is  at 
a  maximum  when  x  =  x0.  Then  as  x,  starting  with  the  value 
x0,  increases,  tan  t,  i.e.  the  slope  of  the  curve,  decreases  alge- 
braically, and  so  the  curve  is  concave  downward  to  the  right  of 
Xq.  On  the  other  hand,  as  x  decreases,  tanr  also  decreases, 
and  so  the  curve  is  concave  upward  to  the  left  of  x0. 

Now,  by  the  above  theorem,  tan  t  will  have  a  maximum  or 
a  minimum  if 

Dx  tanr  =  0,  Dx2  tanr  ^  0. 

Hence,  remembering  that 

t2LTLT  =  Dxyj 

we  obtain  the  following 

Test  for  a  Point  of  Inflection.     If 

[Z>x22/],„=0,  [D'yl^+0, 

the  curve  has  a  point  of  inflection  at  x  =  x0. 


56 


CALCULUS 


This  test,  like  the  foregoing  for  a  maximum  or  a  minimum, 
is  sufficient,  but  not  necessary. 


Then 


Example.     Let 

12y  =  x4  +  2xi-12x2  +  Ux-l. 

12  Dxy  =  4a3  +  6x2  -  24a  + 14, 

12  D2y  =  12  x2  + 12*  -  24  =  12  (x  -  1)  (a?  +  2), 

12Z)x32/  =  12(2a  +  l). 

Setting  ZVy  =  0,  we  get  the  points  x  =  1  and  x  =  —  2.     And 
since 

12[A32/],=.  =  36^0,  l2[2>.'y]„_1=a-36*0) 

we  see  that  both  of  these  points  are  points  of  inflection. 

The   slope   of   the  curve  in  these  points  is  given  by  the 
equations : 

12[A2/]x,i  =  0,  12[D„y]„_i  =  64. 

Hence  the  curve  is  parallel  to  the  axis  of  x  at  the  first  of  these 
points ;  at  the  second  its  slope  is  4i. 


EXERCISES 

Test  the  following  curves  for  maxima,  minima,  and  points 
of  inflection,  and  determine  the  slope  of  the  curve  in  each 
point  of  inflection. 

1.   y  =  4x? -15ar  +  12a?  +  l.       4.   y  =  (x-  l)3(a  +  2)2. 

x 


2.  y  =  x3  +  x4  +  x5. 

3.  6y  =  x6-3x4  +  3x2-l. 


5'   y  =  2  +  3x* 
6.    y  =  (l-x*)s. 

7.  Deduce  a  test  for  dis- 
tinguishing between  two 
such  points  of  inflection 
as  those  indicated  in  Fig. 
12. 


Fig.  12 


APPLICATIONS  57 


8.  On  the  Roots  of  Equations.  The  problem  of  solving  the 
equation 

can  be  formulated  geometrically  as  follows :  To  find  the  points 
of  intersection  of  the  curve 

with  the  axis  of  x : 

2/  =  0. 

Hence  we  see  that  we  can  approximate  to  the  roots  as  closely 
as  we  please  by  plotting  the  curve  with  greater  and  greater 
accuracy  near  the  points  where  it  meets  the  axis  of  x. 

It  is  often  a  matter  of  importance  to  know  how  many  roots 
there  are  in  a  given  interval ;  for  example,  the  number  of  posi- 
tive roots  that  an  equation  possesses.  One  means  of  answer- 
ing this  question  is  by  the  methods  of  curve  tracing  above  set 
forth. 

Consider,  for  example,  the  equation : 

x6-  3^  +  1  =  0. 

The  function 

y=x«-3x2  +  1 

is  positive  for  values  of  x  that  are  numerically  large.     For 


(«-!+!> 


Here  the  parenthesis  apprrAlthes  a  positive  limit  when  x  in- 
creases without  limit ;  the  f~  %  factor  increases  without  limit, 
and  so  the  product  increases  *vi thou t  limit. 

Again,  the  function  has  a  maximum  when  x  =  0  (see  §7),  its 
value  there  being  positive,  namely  1 ;  and  at  x  —  1  it  has  a 
minimum,  its  value  there  being  negative.  Consequently  the 
curve  must  have  crossed  the  positive  axis  of  x  between  x  =  0 
and  x  =  1,  and  again  when  x  >  1,  and  so  the  above  equation 
has  at  least  two  positive  real  roots.     Has  it  more  ? 


58  CALCULUS 

In  the  interval  0  <  x  <  1 

Dxy  =  6x(x*-I)(x>  +  1) 

is  negative  and  hence  the  function  is  steadily  decreasing.    The 

graph,  therefore,  can  have  crossed 
the  axis  of  x  but  once.  Again,  when 
x  >  1,  Dxy  is  positive,  and  so  the 
function  is  always  increasing.  Hence 
x  the  graph  can  have  crossed  the  axis 
h  of  x  beyond  this  point  but  once. 

Thus  we  see  that  the  equation  has 
v!y  just  two  positive  roots,  and  since  to 


FlG  13  each  root  x  =  a  corresponds  a  second 

root  x  =  —  a,  it  has  just  two  nega- 
tive roots,  and  so  in  all  just  four  real  roots. 

A  general  principle  is  illustrated  in  this  example,  which 
may  be  stated  as  follows. 

Theorem.  If  a  continuous  function  f(x)  changes  sign  in 
an  interval  a<x  <  6  and  if  its  derivative  is  positive  at  all  points 
of  the  interval  (or  negative  at  all  points  of  the  interval),  then 
the  function  vanishes  at  just  one  point  of  the  interval. 

The  cubic  equation 

(1)  x3+px  +  q  =  0 

can  be  treated  in  a  similar  manner.     The  graph  of  the  function 

(2)  y  =  Xs  -\-px 

studied  in  §  6  was  especially  s>;  nple.     The  points  in  which 
this  curve  is  cut  by  the  line 

(3)  2/=-? 

evidently  have  for  their  abscissas  the  roots  of  the  cubic  (1).* 
Now  if  p  ^>  0,  the  graph  will  correspond  to  the  first  of  the  two 

*  Another  geometric  formulation  of  the  problem  of  finding  the  roots  of 
the  cubic  (1)  is  to  consider  the  intersections  of  the  curves 

y  =  xs,  y--px-q. 


APPLICATIONS  59 

figures  in  Fig.  10.  Thus  the  line  (3)  will  cut  the  curve  (2) 
in  just  one  point,  and  so  equation  (1)  will  have  just  one  real 
root.*  But  if  p  <  0,  then  the  graph  corresponds  to  the  second 
figure  in  Fig.  10,  and  we  see  that  it  depends  on  the  relative 
magnitudes  of  p  and  q  as  to  whether  there  are  three  points 
of  intersection  or  fewer. 

The  maxima  and  minima  of  the  function  (2)  are  obtained  by 

setting  , 

Dxy  =  3a*+p  =  0,  x=±yJ-£, 

and  it  turns  out  that  a  minimum  occurs  at  the  point  f 


f-V-f'    y=3\-!' 


a  maximum  at 


2  „  =  _?I?J_£. 

3'        y  3\      3 


Hence  if  q  is  numerically  greater  than  these  equal  and  oppo- 
site values  of  y,  i.e.  if  .    s 

y  Z       27' 
the  cubic  (1)  will  have  one  real  root.    If  q,  however,  is  numeri- 
cally smaller :  -    <, 

*  27' 

it  will  have  three  real  roots ;  and  if 

q2  = 


2_      ±PS 


it  will  have  two  real  roots,  one  of  which  counts  twice,  except  in 
the  case  above  mentioned,  p  =  q  =  0,  where  it  has  one  triple  root. 

The  first  is  a  special  cubic,  which  can  be  plotted  accurately  from  the 
tables  once  for  all ;  and  then  the  straight  line  can  be  drawn  as  soon  as 
p  and  q  are  assigned  special  values.  Thus  we  get  a  graphical  solution 
for  any  cubic  of  the  above  type. 

*  In  the  special  case  :  p  —  0,  q  =  0,  the  cubic  (1)  reduces  to  xz  —  0.  It 
is  customary  to  say  that  this  equation  has  three  coincident  roots. 

t  Observe  thatp<0,  so  that  fpV— p/S  is  negative  and,  further  down, 
—  4p3/27  is  positive. 


60  CALCULUS 

We  can  collect  all  cases  under  the  following 
Theorem.     The  cubic  equation 


has 

Xs  +_pa 

'  +  9  =  0 

(a) 

1  real  root  when 

t+ 

27 

>0; 

0) 

3     « 

a         a 

a 

<0; 

(C) 

a         a 

a 

f\     \  P  and  Q  »< 

(<%) 

1        « 

a         « 

a 

«       \   p  =  q=0. 

In  case  (cj)  one  root  counts  twice;  and  in  (c2)  the  root  counts 
three  times. 

EXERCISES 

1.    How  many  real  roots  has  each  of  the  following  equations  ? 

(a)  x5~  5x-  1  =  0.  (c)  3.t4-4x3  +  12.t2  +  7  =  0. 

(6)  3a?4  +  4^  + 6a2- 1  =  0.  (d)  x2n+l  +px  +  a  =  0. 

-4ws.   (a)  three;  (6)  two;   (c)  none. 

2.,  How  many  real  roots  have  the  cubics : 

(a)  a33  +  7a-l  =  0;  (c)   a3-3a-2  =  0; 

(6)  ^-4^  +  1=0;  "(d)  a!3-a  +  3  =  0? 

3.  How  many  positive  roots  has  the  equation 

6xi  +  Sxs-12x2-24:X-l  =  0?  Ans.    One. 

4.  Show  that  the  function 

/(*)  =  (ar-l)8  +  (^2)» 
has  just  one  root  in  the  interval  1  <  x  <  2. 

5.  How  many  real  roots  has  the  equation 

Ix3- 15  x2  +  12a  +  1  =  0?  ^Ins.   Three. 


APPLICATIONS  61 

6.  Show  that,  by  a  suitable  transformation  of  the  coordinate 
axes  to  parallel  axes,  the  new  origin  being  on  the  axis  of  x, 
namely : 

x  =  x'  +  h,  y=y', 

the  equation : 

y  =  x?+p1x2+p2x+p3 

can  be  carried  over  into  the  equation : 

y  =  x*+px  +  q. 

Hence  obtain  the  condition  that  the  cubic: 

x3  +pxx2  +pax  +ps  =  0 

have  three  real  roots. 


CHAPTER   IV 

DIFFERENTIATION  OF  TRANSCENDENTAL  FUNCTIONS 

1.   Differentiation   of  sin  a;.     First  Method.     The   graph  of 
the  function 
(1)  y  =  sin  x 

can  be  constructed  geometrically  by  drawing  a  circle  of  unit 
radius  and  measuring  the  ordinates  corresponding  to  different 


angles ;  the  angle  itself,  measured  in  radians,  giving  the  abscissa 
and  being  computed  arithmetically. 
y 


In  order  to  differentiate  sin  x  we  have  to  give  to  x  an  arbi- 
trary value  x0  and  compute  the  corresponding  value  of  y : 

y0  =  smx0. 

Then  give  x  an  increment,  Ax,  and  compute  again  the  corre- 
sponding value  of  y : 

y0  +  Ay  =  sin  (x0  -f  Ax). 


TRANSCENDENTAL   FUNCTIONS 


63 


Hence 

(2) 


Ay  =  sin  (x0  +  Aa;)  —  sin  #0, 

Ay  _  sin  (#0  -f-  Aa?)  —  sin  x0 

Ax  Ax 


Let  us  follow  these  steps  geometrically  by  constructing  the 
successive  magnitudes.  Fig.  16  explains  itself.  The  radius 
of  the  circle  is  unity,  and  so 

MP  =  sin  xQ,         M'F  =  sin  (x0  +  Ax). 

QP'  =  sin  (x0  -h  Ax)  —  sin  x0  =  Ay,         PP'  =  Ax. 
Hence 

(3)  ^=«Z\ 

Aa;     ppi 

Now  it  is  easy  to  see 
what  limit  this  last  ratio  ap- 
proaches when  P'  approaches 
P.  Suppose  first  we  had  in 
the  denominator  the  chord 
PP'.     Then 

S^=sin<k 
PP1 


M'     M 


Fig.  16 


and  since 

it  follows  that 


lim  <f>  =  ZQPT=?-x0, 
pr=p  2 


lim  -3£—  =  cos  x0. 
p'=  ppp* 


The  chord  of  a  small  angle  differs,  however,  from  the  arc 
only  by  a  small  percentage  of  either.     We  readily  see  that 


W 


r     PP'      1 

lim  — —  =  1. 

pf=p  ppt 


The  student  is  requested  to  draw  an  accurate  figure  represent- 
ing the  arc  and  the  chord  of  an  angle  of  30°  and  also  of  15°. 


64  CALCULUS 

Eeturning  now  to  the  ratio  (3)  and  writing  it  in  the  form : 

QP^^QT    PP 
PP    PP'    pp'' 

we  have,  when  P'  approaches  P  as  its  limit : 

Ax=oAa?      \p'=pppij\p'=pppij 
or  dropping  the  subscripts  : 

(5)  Dx  sin  x  =  cos  x. 

EXERCISE 
Prove  in  a  similar  manner  that 

(6)  Dx  cos  x  =  —  sin  x. 

Second  Method.  The  foregoing  method  has  the  advantage 
of  being  easily  remembered.  Each  analytic  step  is  mirrored 
in  a  simple  geometric  construction.  It  has  the  disadvantage, 
however,  of  incompleteness.  For,  first,  we  have  allowed  Ax, 
in  approaching  0,  to  pass  through  only  positive  values ;  and 
secondly  we  have  assumed  x{)  to  lie  between  0  and  ir/2.  Hence 
there  are  in  all  seven  more  cases  to  consider. 

An  analytic  method  that  is  simple  and  at  the  same  time 
general  is  the  following.  Recall  the  Addition  Theorem  for 
the  sine: 

sin  (a  -j-  b)  =  sin  a  cos  b  +  cos  a  sin  b, 

sin  (a  —  b)  =  sin  a  cos  b  —  cos  a  sin  b, 
whence 

sin  (a  +  b)  —  sin  (a  —  b)  =  2  cos  a  sin  b} 

and  let  a  +  b  =  xQ  +  A#,  a  —  b  =  x*0. 

Solving  these  last  equations  for  a  and  b,  we  get : 

.  Ax  ,      A& 

a  =  Xq  H -,  b  =  —  • 


TRANSCENDENTAL  FUNCTIONS 


65 


Thus         sin  (x0  -f  Ax)  —  sin  x0 


^rVn^> 


Ax" 

2; 


and  the  difference-quotient  (2)  becomes : 


Ay 

Ax 


cos 


.    Ax 
(    ^\      "2" 


The  limit  of  the  first  fac- 
tor on  the  right  is  cos#0. 
The  limit  of  the  second  is 
of  the  form : 

a  =  0       a 

and  here,  again,  we  have  to 

do  with  the  limit  of  the  ratio  of  the  arc  to  the  chord. 

Fig.  17:  ^ 

PP'  =  2sina,  PP'  =  2a. 


For,  in 


We  have  seen  that  the  limit  (4)  has  the  value  1  by  direct 
inspection  of  the  figure.  We  can  give  a  formal  proof  based  on 
the  axioms  of  geometry  as  follows.  First  of  all,  a  straight 
line  is  the  shortest  distance  between  two  points,  and  so 

PP'<PP'. 

Secondly,  if  we  draw  the  tangents  at  P  and  P,  meeting  in  N, 
we  have,  by  the  axiom  that  a  convex  curved  line  is  shorter 
than  a  convex  broken  line  that  envelops  it  and  has  the  same 
extremities :  ^ 

PP'  <  PN+  NP  =  2PN. 

Hence  PP<PP'<2  PN. 

Dividing  through  by  PP'  =  2PM,  and  noticing  that 

PN=     1 

PM     cos  a ' 


we  obtain 


pp 


cos  a 


66 


CALCULUS 


When  a  approaches  0,  1/cosa  approaches  1,  and  thus  the 
variable  PP'/PP'  is  seen  to  lie  between  the  fixed  value  1 
and  a  variable  number  which  is  approaching  1  as  its  limit. 
Consequently  ^p 


lim 


oPP' 


1, 


q.  e.  d. 


The  reciprocal  of  this  ratio,  PP'/PP',  must,  as  was  pointed 
out  in  Chap.  II,  §  5,  under  Theorem  C,  also  approach  unity  as 
its  limit. 


Another  Proof  of  (4). 


The  area  of  the  sector  OAP,  Fig. 
18,  is  \a,  and  it  obviously  lies 
between  the  areas  of  the  tri- 
angles OMP  and  OAQ.    Hence 

^sinctcos  a<  Ja<  Jtana 

or 

.     a       .     1 

cos  a  < < 

sin  a      cos  a 

When  a  approaches  0,  each  of  the  extreme  terms  approaches  1. 
and  so  the  middle  term  must  also  do  so,  q.  e.  d. 
From  Peirce's  Tables,  p.  130,  we  see  that 

sin  4°  40' =  .0814, 

and  the  same  angle,  measured  in  radians,  also  has  the  value 
.0814,  to  three  significant  figures.  Thus  for  values  of  a  not 
exceeding  4° 40',  sin  a  differs  from  a  by  less  than  one  part  in 
800,  or  one-eighth  of  one  per  cent. 

Reason  for  the  Radian.  The  reason  for  measuring  angles  in 
terms  of  the  radian  as  the  unit  now  becomes  clear.  Had  we 
used  the  degree,  the  increment  Ax  would  not  have  been  equal 
to  PP' ;  we  should  have  had  : 


Ax 
360 


PP' 


or 


A        180  $<, 
Ax  = PP'. 


Hence  (3)  would  have  read 


Ay 

Ax 


7T 

180 


QP 
PP' 


TRANSCENDENTAL   FUNCTIONS  67 

and  thus  the  formula  of  differentiation  would  have  become  : 

Dx  sin  x  —  -^-  cos  x. 
180 

The  saving  of  labor  and  the  gain  in  simplicity  in  not  being 
obliged  to  multiply  by  this  constant  each  time  we  differentiate 
is  enormous. 

2.    Differentiation  of  cos  x,  tan  x,  etc.     To  differentiate  cos  x 
we  may  set 


Then     cos  x  =  sin  y, 

Dx  cos  x  =  Dx  sin  y 
=  DJ,smyDxy=—co8y, 

(7)    .  \  Dx  cos  x  =  —  sin  x. 

To  differentiate  tana; 
write 

sin  a; 


tan  x  = 


Then 


COSiC 


Dr  tan  x 


cos2  x  —  sin  x  (  —  sin  x) 


COS^iC 


cos2  a;' 


(8) 


D.  tana; = sec2  x. 


Fig 

19 

» 

J 

J 

£C 

| 

1 

I 

r 

0 

W         /7T 
2    / 

.37T 
2 

Fig.  20 


68  CALCULUS 

EXERCISES 
1.   Show  that 

(a)     Dx  cot  a;  =  —  csc2# ; 

(6)     Dxsecx  —  sin  x  sec2  x; 

(c)     Dx  esc  x  =  —  cos  a;  esc2 a; ;     _- 

(d)*  Dx  vers  a?  =  sin  x. 

Differentiate  the  following  functions : 

•    6.1  —  sin  x.  *  10.   cos3#. 

7.  x  —  tana;.  11.   sec2 a;. 

8.  x  sin  x.  ,  12.    sin  x  cos  x. 

9.    1 .      13.    1~cosx. 

a  sin  a;  +  b  cos  a;  1  +  cos  x 

liml-CO8«  =  0) 

a  =  0  Ot 

first,  by  considering  the  representation  of  numerator  and 
denominator  by  lines  in  Fig.  17  ;  secondly,  by  a  trigonometric 
reduction,  expressing  1  —  cos  a  in  terms  of  the  half-angle,  a/ 2. 

3.   Inverse  Functions.    Let 

be  a  given  function  of  x  and  let  us  solve  for  a?  as  a  function  of 

x  =  <f>(y). 

Then  <f>(y)  is  called  the  inverse  of  the  function  /  (x) .  The 
graph  of  the  former  function  serves  as  the  graph  of  the  latter, 
provided  in  the  latter  case  we  take  y  as  the  independent,  x  as 
the  dependent  variable.     In  order  to  obtain  the  graph  of  the 

*  The  versed  sine  and  the  coversed  sine  are  defined  as  follows  : 
vers  x  —  1  —  cos  x,  covers  x  =  1  —  sin  x. 


<2. 

sin  2x. 

,3. 

cos  2  a;. 

*4. 

,         X 

tan-. 
2 

5. 

sin  a; 

a  +  b  cos  x 

14.   Prove  that 

TRANSCENDENTAL  FUNCTIONS 


69 


inverse  function  with  x  as  the  independent  variable,  transform 
the  x,  y  plane,  and  with  it  the  above  graph,  as  follows :  let 

a  =  2/',        V  =  x'. 

This  is  equivalent  to  rotat- 
ing the  plane  through  180° 
about  the  bisector  of  the 
angle  made  by  the  positive 
coordinate  axes.  In  other 
words,  it  amounts  to  a 
reflection  of  the  plane  in 
that  bisector.  We  have 
met  an  example  of  inverse 
functions  in  the  radical, 
xl'«,  Chap.  II,  §  8. 

If,  as  x  increases,  y 
steadily  increases  (or  if  y 
steadily    decreases),    the 

inverse  function  will  obviously  be  single-valued.     In  this  case 
the  derivative  of  the  inverse  function  is  obtained  from  the 

definition:  , 

y  =  <f>(x)        if        x=f(y), 

and  the  relation : 

Ax     Ax1 
Ay 


Fig.  21 


»*-& 


hence  lim^/=  lim_, 

Ax=oAa;      Ay=oAa? 

Ay 
provided  Dyx  =^0. 

4.   The  Inverse  Trigonometric  Functions,   (a)  sin-1  a.    The 
inverse  of  the  function 
(1)  y  =  sin  x 

is  obtained  as  explained  in  §  3  by  solving  this  equation  for  x 
as  a  function  of  y,  and  is  written  : 

(1')  x  =  sin-1y, 


70 


CALCULUS 


read  "the  antisine  of  y."     In  order  to  obtain  the  graph  of  the 

function 

(2)  y  =  sin-1  x 

we  have,  then,  to  reflect  the  graph  of  (1)  in  the  bisector  of  the 
angle  made  by  the  positive  coordinate  axes.  We  are  thus  led 
to  a  multiple-valued  function,  since  the 
line  x  =  x'(—l<x'<l)  cuts  the  graph 
in  more  than  one  point,  —  in  fact,  in  an 
infinite  number  of  points.  For  most 
purposes  of  the  Calculus,  however,  it  is 
allowable  and  advisable  to  pick  out  just 
one  value  of  the  function  (2),  most  sim- 
ply the  value  that  lies  between  —  tt/2 
and  +7J-/2,  and  to  understand  by  sin"1  a* 
the  single-valued  function  thus  obtained. 
Its  graph  is  the  portion  of  the  curve  in 
Fig.  22  that  is  marked  by  a  heavy  line. 
This  shall  be  our  convention,  then,  in 
the  future  unless  the  contrary  is  ex- 
plicitly stated,  and  thus 


Fig.  22 


(3) 


y  =  sm_Ia; 


is  equivalent  to  the  relations : 
(3')  x  =  sin  y, 

In  particular, 

sin-10  =  0,         sin"1l  = 


< 


Sill 


'(-!)=  -f. 


In  order  to    differentiate  the  function    (3)   write  it  in   the 
implicit  form  (3')  and  differentiate  :  * 

Dxx  =  Dx  sin  y  =  Dy  sin  y  Dxy 

1 


or 


1  =  cos  y  Dxy, 


Dxy  = 


cosy 


*  It  was  shown  in  §  3  generally  that,  whenever  a  function  has  a  deriv- 
ative ^  0,  its  inverse  function  also  has  a  derivative,  and  hence  we  are 


Now 


'RANSCENDENTAL   FUNCTIONS 
or 


71 


sin2?/  +  cos2  y  =  1         or         cos2?/  =  1  —  x2, 

and  since,  for  values  of  y  restricted  as  by  (3')  to  lie  between 
—  tt/2  and  -f  tt/2,    cos?/  >  0,    we  have 


and  so  finally :  * 
(4) 


cos?/=  Vl  —  X2, 

1 


Dxsm~1x  = 


Vl-x2 


(6)  cos  lx.     The  treatment  here  is  precisely  similar.     We 
define : 

(5)  y  —  cos-1  x        if        #  =  cos?/, 

and  we  make  the  inverse  function  single-valued 
by  choosing  that  value  of  y  which  satisfies  the 
relation : 

(6)  0<?/<7r. 
In  particular 


cos-xl  =  0, 


cos-10  = 


2' 


COS"^—  1)=7T. 


(7) 


To  differentiate  cos-1  a?  use  the  implicit  form: 
Dxx  =  Dx  cos  y  =  By  cos  ?/  Z^y, 
1  =  -  sin  y  Dxy, 
1 


Dr cos-1# 


Vl-JB2 


The  functions  sin-1#  and  cos-1ic,  when  restricted  by  (3')  and 
(6)  to  be  single-valued,  are  connected  by  the  relation : 


(8) 


cos-1  a  =  ?  —  sin-1  a?. 


Hence  we  could  have  obtained  (7)  directly  by  differentiating  (8). 

not  assuming  the  existence  of  a  derivative  and  merely  computing  it.  All 
the  conditions  of  Theorem  V  in  Chap.  II  here  employed  are  actually  ful- 
filled, and  thus  our  proof  is  complete. 

*  Geometrically  the  slope  of  the  portion  of  the  graph  in  question  is 
always  positive,  and  so  we  must  use  the  positive  square  root  of  1  —  x2. 


72  CALCULUS 

(c)  tan-1  a;.     Here 

y  =  tan-1  x        if        x  =  tan  y, 

and  we  make  the  inverse  function  single-valued  by  picking  out 
that  value  y  for  which 

do)  ~l<y<l- 


y=\&vrxx 

y 

3* 
2 

^ 

7T 

2 

0 

f                                                   x 

_,__ 

Sir 
2 

(11) 


Fig  24. 
To  differentiate  tan-1  a;  use  the  implicit  form  : 
Dxx  =  Dx  tan  y  =Dy  tan  y  Dxy, 
1  =  sec2yDxy, 
1 


Dxt&TTXX  = 


1  +  x2 


The  other  inverse  trigonometric  functions  are  treated  in 
a  similar  manner.  The  usual  notation  on  the  Continent  for 
sin"*1®,  tan-1ic,  etc.,  is  arc  sin#,  arc  tan  x,  etc. 


TRANSCENDENTAL  FUNCTIONS  73 

Corresponding  to  the  addition  theorems  for  the  trigono- 
metric functions  there  are  functional  relations  for  the  inverse 
trigonometric  functions,  such  for  example  as,  for  tan-1  a; : 

tan_1w  +  tan-1?;  =  tan-1** +  v  • 
1  —  uv 

If  these  relations  are  used,  the  above  definitions  of  sin-1#, 
cos_1#,  tan_1#,  by  which  these  functions  are  made  single-valued, 
must  be  abandoned.  For  this  reason  it  is  better,  for  the 
present,  at  least,  to  abandon  these  relations  and  to  keep  these 
important  functions  single-valued. 


EXERCISES 


1.    Show  that 

(a) 

cot~1a5  =  tan~1-  ; 

X 

(P) 

cot_1ic  =  -  —  tan-1  x. 
2 

2.    Prove  the  following  formulas  : 
1 


(a)     D,  cot_1ic   =  — 


1  +  a2' 


(b)  Dxsec~lx    = .  if      0<  sector    <tt; 

x^/x2  —  1 

(c)  Dxcsc-*x   =       -1      ,  if-^csc-1*    <* 

xVx2  —  !  ^  ^ 

(d)  DxYevs~1x  =  — .  if      0  ^  vers-1  x  ^  ir. 

■y/2x-xl 

Plot  the  graph  of  the  function  roughly  in  each  case. 
3.   Differentiate 

(a)     sin"1;?;  (b)     tan"1-;  (c)     cos-^a. 

2  Q, 


74 


CALCULUS 


5.   Logarithms  and  Exponentials.     In  Chap.  II,  §  8,  we  have 
studied  the  function 

y  =  xn 

for  commensurable  values  of  n,  x  being  the  independent  vari- 
able. Suppose  we  cut  the  family  of  curves  of  Fig.  3  by  a 
straight  line  parallel  to  the  axis  of  ordinates : 

x  =  a,  a>l. 

The  ordinate  of  any  one  of  the  points  of  intersection : 

(1)  y  =  an, 

is  determined  as  soon  as  the  value  of  n  has  been  assigned,  and 
is  thus  a  function  of  n.  From  the  theorem  of  p.  32  we  know 
that,  when  n  increases,  y  increases.  Moreover,  y  is  always  posi- 
tive ;  it  increases  without  limit  when  n  =  -f-  oo ,  and  it  approaches 
0  when  n  =  —  oo . 

As  yet  the  function  (1)  has  been  defined  only  for  commen- 


surable values  of 


What  value  shall  it  have  when,  for  ex- 


ample, n  =  V2?    If  we  allow  n,  passing  through  rational  values, 

to  approach  V2  as 
V  =  e  its  limit,  it  turns  out 

that  an  approaches 
a  definite  limit.  We 
define  a^2  as  this 
limit : 

lim  an  =  a^2. 

n=V2 

And  similarly  for 
every  other  incom- 
mensurable value  of 
the  exponent.  The 
function  thus  ob- 
tained: 
(2)        y=a-, 

is  continuous.  Its 
graph  for  the  special 


TRANSCENDENTAL   FUNCTIONS 


75 


value  a  =  2.72  is  shown  in  Fig.  25.     For  a  proof  of  the  fore- 
going statements  cf.  the  Appendix. 

The  chief  properties  of  the  exponential  function  thus  denned 
are  expressed  by  the  equations 

(I)  au+v  =  auav, 
called  the  Addition  TJieorem;  and 

(II)  (au)v  =  a™. 

The  inverse  of  the  exponential  function  is  the  logarithm : 

(3)  y  =  \ogax        if        x  —  ay. 

It  is  single-valued  and  continuous  for  all  positive  values  of  x. 
Moreover, 

logal  =  0,        logaa  =  l,        logaO  =  -oo,        loga(+co)  =  +  oo. 
V  y  =  \ogx 


Fig.  26 

The  graph  is  obtained  in  the  usual  manner  from 
that  of  (2)  by  reflecting  in  the  bisector  of  the 
positive    coordinate    axes,   and    is    shown   in 
Fig.  26. 
The  chief  properties  of  logarithms  follow  from  (I)  and  (II) : 


(A) 
(B) 


log„a;"  =  nlogaa;. 


76  CALCULUS 

The  proof  of  (A)  is  as  follows.     Let 

u  =  logax,        whence    x  =  au-, 

v  =  logay,  "  y  =  av. 

Then  (I)  becomes: 

au+v  =  xy,        whence     u-\-v  =  \ogaxy, 

or  logaa;  +  log0y  =  log8ajy,  q.  e.  d. 

To  prove  (B)  write  (II)  in  the  form 

(auf  =  anw 

and  substitute  for  au  its  value  a; : 

xn  =  aWM,         whence     tiw  =  loga#n, 

or  loga  a?"  =  n  loga  #,  q.  e.  d. 

A  third  relation  is  of  importance  when  we  have  to  change 
from  one  base  to  another.     It  is : 

(C)  loga*  =  j^. 

It  is  easily  remembered  because  of  its  formal  analogy  with 
the  formula  (V")  of  Chap.  II,  which,  when  the  variables  are 
denoted  by  a,  6,  and  x,  becomes : 

A  « 

The  proof  is  as  follows.     Let 

(a)     w  =  logasc,         whence     x  =  au; 
(/?)     v  =  log6a,  «  x  =  bv; 

(y)    <7=log6a,  "  a  =  bG. 

We  wish  to  prove  that 

v 

From  (y)  follows  that 

b  =  ad. 


TRANSCENDENTAL   FUNCTIONS  77 

Substituting  this  value  of  b  in  (/?)  we  get : 

x  =  ac. 
Substituting  this  value  of  x  in  (a)  we  get : 

au  =  ad. 
And  now  it  merely  remains  to  take  the  logarithm  of  each  side. 
In  particular,  if  we  set  x  =  b,  (C)  becomes : 

(4)  log^^-L.. 

log*  a 

The  following  identity,  which  is  often  useful,  is  obtained  by 
replacing  y  in  the  second  equation  of  (3)  by  its  value  from  the 
first  equation : 

(5)  x  =  a 
Thus 

(6)  xn  —  a 

We  have  assumed  hitherto  that  a  >  1.  If  0  <  a  <  1,  the 
graph  of  (2),  Fig.  25,  must  be  reflected  in  the  axis  of  ordinates, 
and  the  graph  of  (3),  Fig.  26,  in  the  axis  of  abscissas. 


EXERCISES 
1.    Show  that 

(a)  log.-=-log.aj. 

x 

(b)  loga£  =  logaP-logoQ. 


(c)  lo&VT+a^lo&CL  +  a2). 

(d)  loga  (a?  - y2)  =  loga  (x  +  y)  +  loga  (x-y). 

(e)  loga  (x  +  h)-  logax  =  log  A  +  ^  • 
2.  Simplify  the  following  expressions  : 

vi,    V,,    {2>)>,    vpz,    i,    £i;. 


78  CALCULUS 

3.    Solve  the  equation  : 

ax  —  a~x 

=  V 

a*  4.  a-x 

for  x  in  terms  of  y. 


4.    Show  that 


l0*- <%+*>- log,  [(i +*)!]. 


5.  Are  (a)xX        and         (ax)x 
the  same  thing  ? 

6.  If  bx  =  c, 

show  that 

3log06  =  logac. 

6.   Differentiation  of  log  a.     To  differentiate  the  function 
y  =  logax 
we  have  to  form  the  difference-quotient : 

(7)  Ay  =  loga(x0  +  Ax)-logax0==       \       xj 

Ax  Ax  A3. 

and  see  what  limit  it  approaches  when  Ax  approaches  0.     If 
we  set 

Ax 
we  can  write  this  last  expression  in  the  form  : 

3q      A3         \  X0J~XQfl         \  flj"  30      Da|_V  fij    J 

When  A3  approaches  0,  fx  becomes  infinite,  and  the  question 
is  :     What  is  the  value  of  the  limit : 


limfl+i^ 


IV. 


TRANSCENDENTAL  FUNCTIONS  79 

First,  let  /x  become  infinite  passing  through  only  positive 
integral  values : 

fji  =  7l,         ri  =  l,  2,  3,  •••. 
If  we  write 

•M»)-(l+jjT> 

we  get  by  direct  computation : 

*(1)      =2, 

$(2)      =2.25, 

$(3)      =2.37, 

$(10)     =2.59, 

$(100)    =2.70, 

$(1000)   =2.72, 

$(10,000)  =2.72. 

Hence  we  see  that,  as  n  increases,  <f>(ri)  increases,  but  does 
not  appear  to  mount  above  a  number  somewhat  less  than  3. 
We  can  show  this  to  be  the  case  no  matter  how  large  n.  By 
the  Binomial  Theorem : 

(a  +  b)n  =  an  +  nan~l b  +  n(n~1)an-2b2  + ... . 

JL  •  £ 

we  have : 


n(n-l)(n-2)  /l V 
1-2.3        Uy 

-  Mf1--) 

n      \        7iy  \        71/ 


H to  n  + 1  terms 


1- 

=1  +  1  +  T-|+V      1^2. 3    'V  + 

These  terms  are  all  positive  and  each  increases  when  n  in- 
creases. Moreover,  when  n  increases,  additional  positive  terms 
present  themselves.     And  so,  for  both  reasons,  $(71)  increases: 

$(tt  +  l)>$(7l). 


80  CALCULUS 

Secondly,  <{>(n)  is  always  less  than  3.  For,  the  above  terms 
after  the  first  two  are  less  than  the  terms  of  the  series: 

1  +  1  +  1  +  1+  ...  +_!_. 
^    T  2     22  2n~x 

Kecalling  the  formula  for  the  sum  of  the  first  n  terms  of  a 
geometric  progression,  Chap.  I,  §  1,  and  setting  a  =  1,  r  =  J, 
we  get: 

1  +  1+1+  ...+_L  =  lz^i)!  =  2— 1, 

and  so:  <f> (/i) <3—  -— -<  3. 

r  v  '  2n 

Now  if  we  have  a  function  of  the  positive  integer  n  which 
always  increases  as  n  increases,  but  never  exceeds  a  certain 

-h 1 "i?r\  ml  i 

0  1  2  '      3 

Fig.  27 

fixed  number,  A,  it  must  approach  a  limit  not  greater  than  A, 
when  n  =  oo  (Fundamental  Principle  for  the  existence  of  a 
limit,  Chap.  XII,  §  2) .  Hence  <f>  (n)  approaches  a  limit  whose 
value,  e,  is  not  greater  than  3 : 

(9)  lin/l+iY=e. 

„=»  ^        nj 

We  shall  see  later  that 

e  =  2.718.-.. 

/u.  irrational  and  negative.  We  can  now  show  that,  when  fx 
becomes  positively  infinite,  varying  continuously,  i.e.  passing 
through  all  positive  values,  <£(/u.)  still  approaches  e  as  its  limit. 

Let  n</*<w  +  l, 


where  n  is  an  integer ;  then 


71  +  1  //,  W 

fr^r<^i)'<('^y. 


TRANSCENDENTAL   FUNCTIONS  81 

We  shall  only  strengthen  this  inequality  if  in  the  left-hand 
member  we  replace  ll  by  n,  in  the  right-hand  member,  by  n-fl. 
Hence 

n+1 

As  ix,  and  with  it  n,  becomes  infinite,  each  extreme  member  of 
this  double  inequality  approaches  e  as  its  limit,  hence  the 
mean  member  must  likewise  approach  e  and 

lim  <f>  (ji)  =  e. 

M  =  +  «> 

Finally,  let  p  be  negative,  p  =  —  r.     Then 

K)-=K)"H^)H-^H1+^J 

When  fi  =  —  oo  ,  r  =  +00  ,  and 

lim  <£(/*)  =  e. 

/A  =  — 00 

Returning  now  to  equations  (7)  and  (8)  and  remembering 
that  loga#  is  a  continuous  function,  we  see  that 


&y 


lim^=  lim  rilog/l  +  iYl  =  ilogriim  fl  +  -Y' 
=  -logae, 
or,  dropping  the  subscript,  we  have : 

Dx\ogax  =  1-^. 
x 


82  CALCULUS 

The  base  a  of  the  system  of  logarithms  which  we  will  use  is 
at  our  disposal.     We  may  choose  it  so  that  the  constant  factor 

log„e  =  1, 

namely,  by  taking  e  as  the  base  :  a  =  e.     Thus  (10)  becomes : 

(11)  D.log>=i. 

This  base,  e,  is  called  the  natural  base,  and  logarithms  taken 
with  e  as  the  base  are  called  natural  or  Napierian  logarithms,  in 
distinction  from  denary  or  Briggs's  logarithms,  for  which  the 
base  is  10.  Natural  logarithms  are  used  in  the  Calculus  be- 
cause of  the  gain  in  simplicity  in  the  formulas  of  differentia- 
tion and  integration, — a  gain  precisely  analogous  to  that  in 
the  differentiation  of  the  trigonometric  functions  when  the 
angle  is  measured  in  radians. 

It  is  customary  in  the  Calculus  not  to  write  the  subscript  e 
and  to  understand  by  log  a;  the  natural  logarithm  of  x,  the 
denary  logarithm  being  expressed  as  log10#. 

7.  The  Compound  Interest  Law.  The  limit  (9)  of  §  6  pre- 
sents itself  in  a  variety  of  problems,  typical  for  which  is  that 
of  finding  how  much  interest  a  given  sum  of  money  would  bear 
if  the  interest  were  compounded  continuously,  so  that  there  is 
no  loss  whatever.  For  example,  $1000,  put  at  interest  at  6  %, 
amounts  in  a  year  to  $1060,  if  the  interest  is  not  com- 
pounded at  all.  If  it  is  compounded  every  six  months,  we 
have 


$1000 


(1  +  tt) 


as  the  amount  at  the  end  of  the  first  six  months,  and  this  must 
be  multiplied  by  (1  +  '-— -  J  to  yield  the  amount  at  the  end  of 
the  second  six  months,  the  final  amount  thus  being 


$1000 


(>+3)f 


TRANSCENDENTAL  FUNCTIONS  83 

It  is  readily  seen  that  if  the  interest  is  compounded  n  times 
in  a  year,  the  principal  and  interest  at  the  end  of  the  year  will 
amount  to 

1000(1 +^)* 

dollars,  and  we  wish  to  find  the  limit  of  this  expression  when 
n  =  00  .     To  do  so,  write  it  in  the  form  : 


1000 


(i+^ 


n  J 


and  set  n/.06  =  fi.     The  bracket  thus  becomes 

♦w-(i  +  iy: 

and  its  limit  is  e.     Hence  the  desired  result  is 
1000  e06=  1061.84* 

EXERCISE 

If  $  1000  is  put  at  interest  at  4  %  compare  the  amounts  of 
principal  and  interest  at  the  end  of  10  years,  (a)  when  the 
interest  is  compounded  semi-annually,  and  (&)  when  it  is  com- 
pounded continuously.  Ans.   A  difference  of  -$5.88. 

8.  Differentiation  of  e*,  ax.  Since  ax  and  logax  are  inverse 
functions,  we  have : 

(12)  y=ax        if        x  =  \ogay. 
In  particular  : 

(13)  y=e*  if        x  =  \ogy. 

Differentiating  this  last  equation  with  respect  to  x,  we 
obtain : 

Dxx  =  Dx\ogy  =  DylogyDxy,  1  =  \DxVy 

(14)  .-.     Dxex  =  (f. 

*The  actual  computation  here  is  expeditiously  done  by  means  of 
series ;  see  the  chapter  on  Taylor's  Theorem. 


84  CALCULUS 

If  we  proceed  with  (12)  in  a  similar  manner,  we  get  as  our 
first  result : 

y 

By  (4)  in  §  5 : 

logae  = 


logea 
(15)  .*.    Dxax  =  a*  log  a. 

Differentiation  ofxn,  n  irrational.  Formula  (12)  of  Chap.  II 
can  now  be  shown  to  hold  when  n  is  irrational.  Since  by 
§5,(6): 

if  we  set  z  —  n\ogx, 

we  have      Dxxn  =  Dxe*  =  D*e"  •  Dxz=e'n-  =  nxn~1,  q.  e.  d. 

x 

We  are  now  in  a  position  to  differentiate  any  of  the  elemen- 
tary functions  without  evaluating  new  limits,  for  any  such 
differentiations  can  be  reduced,  by  the  aid  of  Theorems  I-V  of 
Chap.  II,  to  special  formulas  already  in  our  possession.  An 
important  aid,  however,  in  the  technique  of  differentiation  is 
furnished  by  the  method  of  differentials,  which  we  will  con- 
sider in  the  next  chapter,  and  so  we  shall  postpone  the  drill 
work  on  this  chapter  till  that  method  has  been  taken  up. 


EXERCISES 

Differentiate  the  following  functions : 

1.  y  =  log10oj.  Ans.   Dxp  = 

x 

2.  y  =  10x.  Ans.   Dxy  =  2.303  x  l(f 

3.  2/  =  log  sin  x.  4.    y  =  eC08x. 

5.   2/  =  k)g  tan (^+2^  Ans.  D9y=  — 

\J2      4/-  cos  x 


CHAPTER  V 

INFINITESIMALS  AND  DIFFERENTIALS 

1.  Infinitesimals.  An  infinitesimal  is  a  variable  which  it  is 
usually  desirable  to  consider  only  for  values  numerically  small 
and  which,  when  the  formulation  of  the  problem  in  hand  has 
progressed  to  a  certain  stage,  is  allowed  to  approach  0  as  its 
limit.  Thus  in  the  problem  of  differentiation  Ax  and  Ay  are 
infinitesimals  ;  for  we  allow  Aa;  to  approach  0  as  its  limit  and 
then  Ay  in  general  also  approaches  0.     Again,  in  Chap.  IV,  we 

had  to  do  with  lim- — ^-   Here  a  and  sin  a  are  infinitesimals. 

a=0      a 

Further  examples  of  infinitesimals  are  furnished  by  the  fol- 
lowing magnitudes  of  Fig.  28  : 

(1)     a  =  AP,         ft=AQ  =  tsma,  t 


y  =  MA  =  l-cosa,         8  =  AJST,  0  ^<  K 

e=PQ,        Z  =  AQ+QP,     etc.  Fig.  28. 

That  infinitesimal  which  is  chosen  as  the  independent  vari- 
able is  called  the  principal  infinitesimal. 

Two  infinitesimals,  a  and  ft,  are  said  to  be  of  the  same  order 
if  their  ratio  approaches  a  limit  not  0  when  the  principal  infini- 
tesimal approaches  0 : 

lim  £=  K^O. 
a 

If  their  ratio  approaches  0  as  its  limit,  ft  is  said  to  be  of 
higher  order  than  a,  and  if  it  becomes  infinite,  ft  is  of  lower 
order. 

85 


86  CALCULUS 

Thus  if  /3  =  tana, 


ft  __  tan  a  _     1 

since 

a         a         cos  a 

a 

Ld                                     lim£  =  l=£0. 

a  =  0  a 

ence  tan  a  is  of  the  same  order  as  a. 

Again,  if                         y  =  1  —  cos  a, 

2sin2£ 
y      1  —  cos  a               2 

sin£ 

•    a       2 

sin-. 

2    a 

2 

ence                                     lim  1  =  0 

a=0C6 

and  1  —  cos  a  is  of  higher  order  than  a. 

An  example  of  an  infinitesimal  of  lower  order  is  V«.     For 

V«       1  A         r      1 

=  — -        and         lim  — =  =  oo 

a         V^  a=0  ^/a 

(read  :  "1/Va  becomes  infinite  when  a  approaches  0"). 

It  is  readily  seen  that,  if  (3  and  y  are  infinitesima  s  of  the 
same  order,  or  if  y  is  of  higher  order  than  j3,  and  if  further- 
more p  is  of  higher  order  than  a,  then  y  is  also  of  higher  order 
than  a.     For,  since 

lim  1=K. 
we  have:  1  =  k+€,        y=#/3  +  eft 

where  c  is  infinitesimal.     Hence 

a      ct  J 

and  since  lim  p/a  =  0  by  hypothesis,  we  see  that  lim  y/a  =  0. 


INFINITESIMALS  AND   DIFFERENTIALS 


87 


Similarly,  if  ft  and  y  are  of  the  same  order,  or  if  y  is  of 
lower  order  than  /?,  and  if  furthermore  /?  is  of  lower  order 
than  «,  then  y  is  also  of  lower  order  than  a. 

An  infinitesimal  y3  is  said  to  be  of  the  n-th  order  if  p/an 
approaches  a  limit  not  0,  a  being  the  principal  infinitesimal : 

lim  £  =  K^  0. 

a=oan 

For  example,  if  a  is  the  principal  infinitesimal,  sin  a  and 
tan  a  are  of  the  first  order,  Va  is  of  order  J-,  an  is  of  order  n, 
and  1  —  cos  a  can  be  shown  to  be  of  order  2.     For 


2  sin2? 


COSrt 


sin- 


and 


lim 

a=b0 


cos  a 


Theorem.    If  two  infinitesimals  a  and  fi  differ  from  each  other 
by  an  infinitesimal  of  higher  order  than  either,  then 


P 


lim  tL 
a 


I 


And  conversely:    If limp  /a  =  1,  then  a  and  /?  differ  from  each 
other  by  an  infinitesimal  of  higher  order  than  either : 


/?-«= 


lim 


0. 


First,  our  hypothesis  is  that,  if  we  write 
^8  —  a  =  c,         then 
Dividing  through  by  a  we  have : 


lim  1  =  0. 
a 


Hence 


lim 


|8_ 


lim 


H-. 


q.  e.  d. 


88  CALCULUS 

To  prove  the  converse,  write 

£=!  +  „. 
a 

Then  rj  is  infinitesimal.     Multiplying  up  we  have : 

P  =  a  4-  ya> 

The  difference,  ft—  a  =  r)a  =  e,  is  evidently  an  infinitesimal  of 
higher  order  than  a  and  hence  also  than  /?. 

Definition.     If  a  is  the  principal  infinitesimal  and  if 

lim  fi  _  jT- 


a=0 


so  that,  when  we  write 


£=ir+£) 


a 

we  have  (3  =  Ka  +  ea, 

then  the  term  Ka  is  called  the  principal  part  of  p. 

EXERCISES 

^  1.    Show  that  a  —  2  a2  and  3  a  -f  a?  are  infinitesimals  of  the 
same  order. 

1  2.   Show  that  ot—  sin  a  is  of  higher  order  than  a. 

« 3.   Show  that  asin«  is  an  infinitesimal  of  the  second  order. 

'  4.    In  Fig.  28  show  that  PQ  and  MA  are  infinitesimals  of 
the  same  order. 

5.  Determine  the  order  of  AR,  referred  to  a. 

)  6.  Show  that  AN  is  of  higher  order  than  RQ. 

7.  Show  that  AQ  and  MP  are  of  the  same  order. 

8.  Show  that  PQ  is  of  the  second  order,  referred  to  a. 

9.  Determine  the  order  of  each  of  the  following  infini- 
tesimals : 


(a)  a  +  sin  a;         (b)  Vsina;         (c)  VI  —  cos  a- 


INFINITESIMALS   AND   DIFFERENTIALS  89 

10.  Show  that  the  sum  of  two  positive  infinitesimals,  each 
of  the  first  order,  is  always  an  infinitesimal  of  the  first  order, 
and  that  the  difference  is  never  of  lower  order.  Cite  an  example 
to  show  that  the  difference  may  be  of  higher  order. 

11.  Determine  the  principal  part  of  each  of  the  infinitesi- 
mals in  the  text  numbered  (1). 

12.  If  two  infinitesimals  have  the  same  principal  part,  show 
that  they  differ  from  each  other  by  a  small  percentage  of  the 
value  of  either,  and  that  this  percentage  is  infinitesimal,  pro- 
vided that  their  principal  part  is  not  0. 

# 

2.  Fundamental  Theorem.  There  are  two  theorems  that  are 
fundamental  relating  to  the  replacement  of  infinitesimals  by 
other  infinitesimals  that  differ  from  them  respectively  by 
infinitesimals  of  higher  order.  One  is  the  theorem  of  this 
paragraph ;  the  other  is  Duhamel's  Theorem  of  Chap.  IX,  §  6. 

Theorem.  In  taking  the  limit  of  the  ratio  of  tivo  infinitesi- 
mals, each  infinitesimal  may  be  replaced  by  another  one  which 
differs  from  it  by  an  infinitesimal  of  higher  order : 

lim£  =  lim£' 
y  7 

if  lim^  =  l         and         lim^  =  l. 

P  y 


For,  since 


|'  =  l  +  €         or         £'  =  0(1 +  c) 
P 


and  -==1  +  v      or       y'  =  y(1  +  >?)> 

where  c  and  rj  are  infinitesimals,  we  have : 

y'    y!  +  v 

Hence 

lim  #  =  Aim  £Ylim  ±±±)  =  lim  &,  q. e.  d. 

y'    \      yA      i  +  W  y 


90 


CALCULUS 


3.  Tangents  in  Polar  Coordinates.    Let 

be  the  equation  of  a  curve  in  polar  coordinates.  We  wish  to 
find  the  direction  of  its  tangent.  The  direction  will  be  known 
if  we  can  determine  the  angle  \jj  between  the  radius  vector 
produced   and    the    tangent.      Let  P,   with   the   coordinates 

(r0,  00),  be  an  arbitrary 
point  of  the  curve  and 
P':(r0  +  Ar,  00  +  A0) 
a  neighboring  point. 
Draw  the  chord  PP 
and  denote theZOP'P 
by«//.  Then  obviously 

lim  «//=  xjjq. 


Fig.  29 


To    determine    \f/0j 
drop  a  perpendicular 
PM  from  P  on  the  radius  vector  OP  and  draw  an  arc  PN  of 
a  circle  with  0  as  centre.     The  right  triangle  MP'P  is  a  tri- 
angle of  reference  for  the  angle  if/'  and 


tani// 


MP 
PM 


Hence 


tan  \pQ  =  lim  tan  «//=  lim 


MP 


p±pP'M 


In  the  latter  ratio  we  can,  by  virtue  of  the  Fundamental 
Theorem  of  §  2,  replace  MP  and  P'M  by  more  convenient 
infinitesimals.     We  observe  that 


MP=rosinA0, 
Furthermore, 

PN=  Ar, 


hence 


so  that 


,.      MP 
lim 

A9=orf,A0 


1. 


lim 


P'M 


1. 


A0=o   Ar 


INFINITESIMALS  AND  DIFFERENTIALS  91 

Hence  we  have :  \  /> 

lim#£=lim^  =  [ri^_,, 
p±pPM      m*o  Ar       L  J<M,0, 

or,  dropping  the  subscripts : 

(2)  tani/r  =  rDr0. 

Example.     The  curve 

(3)  r  =  ae™,  a>0, 

is  a  spiral,  except  when  A.  =  0,  which  coils  round  the  origin 
infinitely  often.     Here, 

Der  =  a\eK»,  Dr0  =  -?-  =  — -,     tan^  =  ~,  or  cotxj/  =  \. 
D9r     a\eK9  \ 

Hence  the  tangent  always  makes  the  same  angle,  cot-1  A.,  with 
the  radius  vector  produced.  For  this  reason  the  curve  is  called 
the  equiangular  spiral. 

EXERCISES 

1.  Plot  the  curve 

r  =  6, 

and  determine  the  angle  at  which  it  crosses  the  prime  vector 
when  r  =  2tt.  Ans.   \f/  =  81°,  nearly. 

2.  The  equation  of  a  parabola  referred  to  its  focus  as  pole  is 

r  (1  +  cos  0)  =  m. 
Find  the  value  of  if/  when  0  =  0  and  when  0  =  tt/2. 

3.  The  equation  of  a  cardioid  is 

r=a(l  —  cos0). 
Determine  \f/. 

4.  Differentials.     Let 

be  a  function  of  a;  and  let  Dxy  be  its  derivative : 


Km  *H  =  Dxy. 

Ax=oAa; 


~) 


92 


CALCULUS 


Then 


Ay 

Ax 


Dxy  +  t, 


where  c  is  infinitesimal,  and 

Ay  =  Dxy  Ax  +  e  Ax. 

Since  x  is  the  independent  variable,  we  may  take  Ax  as  the 
principal  infinitesimal,  and  this  last  relation  represents  Ay  as 
the  sum  of  its  principal  part,  DxyAx,  and  an  infinitesimal  of 
higher  order,  e  Ax, 

Definition.     The  principal  part  of  the  increment  Ay  of  the 
function  (1)  is  called  the  differential  of  y  and  is  denoted  by  dy : 

(2)  dy  =  DxyAx. 

We  may,  in  particular,  choose  f(x)  as  the  function  x.     Then 

(2)  becomes  : 

(3)  dx=DxxAx  =  Ax. 

Thus  we  see  that  the  differential  of  the  independent  variable,  x, 
is  equal  to  the  increment  of  that  variable.  But  this  is  not  in 
general  true  of  the  dependent  variable,  since  e  does  not  in 
general  vanish. 

By  means  of  (3)  equation  (2)  can  be  written  in  the  form  : 


dy  =  Dxydx. 


D*y. 


x' 
Fig.  30 


Geometrically,  the  in- 
crement Ay  of  the  func- 
tion is  represented  by  the 
line  MP'  in  Fig.  30,  while 
the  differential,  dy,  is  equal 
to  MQ.  It  is  obvious 
geometrically  that  the  dif- 
ference between  Ay  and 
dy,  namely  the  line  QP1, 
is   of    higher   order    than 


INFINITESIMALS  AND  DIFFERENTIALS  93 

Ax  —  PM.     The  triangle  PMQ  is  a  triangle  of  reference  for  t, 

and  .  dy 

tan  t  =  -^  • 
dx 

In  the  above  definition  x  has  been  taken  as  the  independent 
variable,  Ax  as  the  principal  infinitesimal.  The  following 
theorem  is  fundamental  in  the  theory  of  differentials. 

Theorem.     The  relation  (4)  : 

dy  =  Dxydx, 

is  true,  even  when  x  and  y  are  both  dependent  on  a  third  vari- 
able, t. 

Suppose,  namely,  that  x  and  y  come  to  us  as  functions  of  a 
third  variable,  t : 

(6)  x  =  <f>(t),  y  =  t(t), 

and  that,  when  we  eliminate  t  between  these  two  equations,  we 
obtain  the  function  (1).  Then  dx  and  dy  have  the  following 
values,  in  accordance  with  the  above  definition,  since  t,  not  x, 
is  now  the  independent  variable,  At  the  principal  infinitesimal : 

dy  =  Dty  At,  dx  =  DtxAt. 

We  wish  to  prove  that 

dy  =  Dxydx. 

Now  by  Theorem  V  of  Chap.  II : 

Dty  =  DxyDtx. 

Hence,  multiplying  through  by  A£,  we  get : 

DtyAt  =  Dxy.DtxAt, 

or  dy  =  Dxydx,  q.  e.  d. 

With  this  theorem  the  explicit  use  of  Theorem  V  in  Chap. 
II  disappears,  Formula  V  of  that  theorem  now  taking  on  the 
form  of  an  algebraic  identity  : 

du  _  du  dy 
dx      dy  dx 

To  this  fact  is  due  the  chief  advantage  of  differentials  in  the 
technique  of  differentiation. 


94  CALCULUS 

Differentials  of  Higher   Order.     It  is  possible  to  introduce 
differentials  of  higher  order  by  a  similar  definition  : 

d2y  =  DJyAx2,  d3y  =  D^yAx3,     etc. 

But  inasmuch  as  a  theorem  analogous  to  the  above  for  differen- 
tials of  the  first  order  does  not  hold  true  here,  the  chief  advan- 
tage of  the  differentials  of  the  first  order  is  lost.  We  shall, 
therefore,  refrain  from  introducing  differentials  of  higher  order 
and  regard  the  expressions  : 

%  %  etc, 

dor  dx3 

not  as  ratios,  but  merely  as  another  notation  for  the  deriva- 
tives D2y,  Dxsy,  etc.* 


Remark.     The  operator  Dx  is  written  in  differential  form  as 
dx 


Thus 


t-^t1-      means      ^Vr^-' 

dx*l  —  x  *  1  —  x 

and  similarly  for  higher  derivatives. 


5.   Technique  of  Differentiation.     Theorems  I-IV  of  Chap. 

II,  written  in  terms  of  differentials,  are  as  follows. 

General  Formulas  of  Differentiation 

I.  d  (cu)  =  c  du. 

II.  d  (u  +  v)  =  du  +  dv. 

III.  d  (uv)  =udv+  v  du. 
v  , / 'a\  _vdu  —  udv 


vj  v2 

*  Differentials  of  higher  order  are  still  used  in  some  branches  of  mathe- 
matics, notably  in  differential  geometry.  For  a  treatment  of  such  differ- 
entials cf.  Goursat-Hedrick,  Mathematical  Analysis,  vol.  I,  §  14. 


INFINITESIMALS   AND   DIFFERENTIALS  95 

Consider,  for  example,  the  first : 

rw    \         ,->  .  d(cu)        du 

Dx(cu)  =  cDxu,        i.e.         — * — t-  =  c — , 

dx  dx 

and  it  remains  only  to  multiply  through  by  dx.  —  We  have 
already  noted  the  disappearance  of  Theorem  V. 

To  these  are  to  be  added  the  special  formulas  of  Chapters  II 
and  IV.  Beside  the  derivatives  that  were  there  worked  out 
ab  initio  it  is  useful  to  include  in  this  list  a  few  others. 


Special  Formulas  of  Differentiation 


1. 

dc 

=  0. 

2. 

dxn 

=  nxn~1dx. 

3. 

dsinx 

=  cos  x  dx. 

4. 

dcosx 

=  —  sin  x  dx. 

5. 

d  tan  x 

=  sec2xdx. 

6. 

dlogx 

_dx 

X 

7. 

dex 

=  exdx. 

8. 

c?sin_1a; 

dx 

9.  dtsai^x 


VI  -x2 

dx 

1+a*' 


This  list  is  somewhat  elastic.     Some  students  will  prefer  to 
include  the  formulas  : 

10.  dcos~lx     t= 


11.  dvers-1#  = 


Vl-x* 
dx 


V2X-X2 
12.  da*  =axlogadx. 

But  it  is  better  to  err  on  the  side  of  brevity. 


96  CALCULUS 

All  other  functions  that  occur  as  combinations  of  the  above,  — 
the  so-called  elementary  functions,  —  can  be  differentiated  by 
the  aid  of  these  two  sets  of  formulas.  We  will  illustrate  the 
use  of  differentials  by  some  examples. 

Example  1.   To  differentiate 

y—  Va^  —  x2. 


Let 

z  —  a2  —  x2. 

Then 

y  =  z\ 

dy=  dz*  =  ±z~*dz  =\z~%  (—  2xdx), 

dy  _      2          x 

ax          3</(a2-a*Y 

The  work  can,  however,  be  abbreviated  and  rendered  more 
concise  by  refraining  from  writing  a  new  letter,  z,  for  the 
function  a2  —  x2 : 

dy=  \{a2  -  x2y%d(a2  -x2)  =  -f  a(a2-  x^dx. 

Example  2.     To  differentiate 

y  =  log  sin  x. 

Let  z  —  sin  x,  y  =  log  z. 

rn,                                j         ii             dz     cosxdx 
Then  dy  =  d  log  2  =  —  = , 

z  z 

dy     cos  x  , 

or  —  = =  cot  x. 

dx     sin  x 

More  concisely : 

,         ,,  d  sina?     cosae&c 

dy  =  d  log  sin  a;  =  — : =  — : • 

sm  x         sm  x 

Example  3.     To  differentiate 

u  =  eot  cos  6?. 


INFINITESIMALS   AND   DIFFERENTIALS  97 

du  sb  cos  bt  deat  +  eM  dcos  bt 

=  cos  bt  e°*  d(at)  —  eM  sin  bt  d(bt) 

ss  ea'(acos  bt—  b  sin  6£)  d£, 

—  =  eat  (a  cos  bt  —  b  sin  &£). 
eft  y 

When  the  student  has  had  some  practice  in  the  use  of  differ- 
entials he  will  have  no  difficulty  in  suppressing  the  first  two 
lines  of  this  last  differentiation. 

Example  4.     To  differentiate  y,  where 

Xs  —  3  xy  +  y4  =  1. 

dx3-3d(xy)+dy4  =  dlf 

3x2dx  —  3xdy  —  3ydx  +  4:ysdy  =  0, 

dy  =  3x2-3y 
dx      3x  —  4:ys 

The  student  will  avoid  errors  by  noting  that  when  one  term 
of  an  equation  is  multiplied  by  a  differential,  every  term  must 
be  so  multiplied.  Thus  such  an  equation  as  dy  =  x2  —  3  x  is 
impossible. 

EXERCISES 

Employ  the  method  of  differentials  for  performing  the  fol- 
lowing differentiations. 

du  =    (b  +  2cx)dx    . 
2  Va +•&#-[-  ex2 

x2—  1 

dy = — i—  dx- 

XT 

ds  -3 


1. 

u  = 

=  Vc 

i  +  bx  +  ex2. 

2. 

y  = 

1- 

-2x  +  x* 

X 

3. 

s  = 

_  1- 

-t 

1+2*  dt      (1  +  2*)2 

4.    y  =  (l-x)(2-3x)(5-2x). 
Work  in  two  ways  and  check  the  answers. 


98  CALCULUS 

5.  r  —  aek6.  9.  y  =  logcosx. 

6.  7/  =  e-'(2^4-6^2-3^-3).        10.  y  =  log(ex-  e~x). 

7.  u  =  ^¥=rfV^x~.  11.  y^Bing+ooBs, 

%  8.    y  =  log  a  "*~ g  ■  12.  w  =  log  Vl  —  cos  x. 

a  —  x 

13.    #  =  Vl  +  sinv.  ~  = —  ■ 

dx      V2-Z2 


*      ~     1-ar2 

dx     1+x2 

15. 

y  =  tan-1    x+    . 

vl7. 

w  =  cot  x  -  • 

16. 

2/  =  sin-1  (n  sin  x). 

18. 

u  =  cos  *  - • 

*    19. 

2,  =  logtan(|+|). 
If     y  =  a?emx,     find 

dy 

-3-  =  sec  x. 

dx 

20. 

d2y 

dx2 

and 

dsy 

dx5 

21.  The  same  when  y  =  x2emx. 

22.  Differentiate 

23.  Differentiate : 


22.   Differentiate     x2ax.  — x2ax=2xax +  x2axloga. 

dx 


(a)     xKF;  (b)     10*2;  (c)     afw*. 

24.  Find  the  slope  of  the  curve  y  =  log10x  at  the  point  x  =  1, 
?/  =  0.  .4ws.   tan  t  =  .4343. 

25.  Obtain  the  equations  of  the  tangent  and  normal  of  the 
curve  y  =  10x  at  the  point  where  it  crosses  the  axis  of  ordinates. 

To  differentiate  a  function  of  the  form 

begin  by  taking  the  logarithm  of  each  side  of  the  equation : 
\ogy  =  <f>(x)logf(x). 


INFINITESIMALS   AND   DIFFERENTIALS  99 

Or,  what  amounts  to  the  same  thing,  write 

f(x)  =  elog/(x),         \_fix)'V>{x)  =  eWx)l0*'(x). 
Thus,  to  differentiate  y  =  xx,  write 

log  y  —  x  logic  or        y  —  exloex. 

Hence     -^=d(x  log  x)  —  etc.      or     dy  —  ex  log  x  d  (x  log  x)  =  etc. 

y 

g=af(l+log*). 

Differentiate  each  of  the  following  functions : 

i 

26.  y  =  xx.  28.    2/  =  (cos#)tan*. 

27.  y  =  x*[nx.  29.    ?/  =  (sin  x)'inx. 

6.  Differential  of  Arc.  Let  s  denote  the  length  of  the  arc 
of  the  curve  y  =/(#),  measured  from  a  fixed  point  A,  and  let 
As  denote  the  length  of  the  arc  PP.     Then  (see  Fig.  30) 

PP'2  =  Aa2  +  Ay2, 

hence  lim  (*£)*  =  1  +  lim  /^Y. 

Ax=o  \  A# /  Ax=o\^.xJ 

In  taking  the  first  limit  we  can  replace  the  chord  PP'  by  the 
arc  As,  since  * 

lim——  =  1. 


pf=p 


PP 


*  A  formal  proof  of  this  statement  can  be  given  as  follows.    We  have : 

(A)  PP'<PP'<PQ+QP', 

since  (a)  a  straight  line  is  the  shortest  distance  between  two  points  and 
(b)  a  convex  curved  line  is  less  than  a  broken  line  that  envelops  it  and 
has  the  same  extremities.     From  (A)  follows  : 

pp,    pp,    pp, 


100  CALCULUS 

We  thus  obtain  the  relation : 

(1)  {D98)*  =  l+(Dxy)*        or         ds2  =  dx2  +  dy\ 

Geometrically  we  see  that  ds  is  represented*  by  the  hypothe- 
nuse  PQ  of  the  right  triangle  PMQ.     Furthermore  : 


(2) 


SlllT 


COST: 


dy 


d*      -y/oW+dy 

dx  dx 

<&  ~  Vdx2  +  dy2 


dy 
dx 


■  v 


1  + 


df 
tx~2 


4 


1  + 


dy2 
dx2 


These  formulas  are  written  on  the  supposition  that  the  tan- 
gent FT  is  drawn  in  the  direction  in  which  s  increases  and  that 
x  and  s  increase  simultaneously.  If  x  decreases  as  s  increases, 
we  must  place  a  minus  sign  before  each  radical. 

Polar  Coordinates.  Similar  considerations  in  the  case  of  the 
curve  r=f(6)  lead  to  the  following  formulas,  cf.  Fig.  29  : 

PP"=  PM>  +  MP>,      lim  (&)*-  lim  (£*Y  +  ]im  (ML\  '. 
A/-=o\  Ar  J       Ar=o\  Ar  J       A/-=o^  Ar  J 


(3) 


(Drs)2  =  1  +  i*(Drff)*        or        ds2  =  dr2  +  rW. 


As  P'  approaches  P,  the  ratio 


*        dx* 


PQ  _  Vdx*  +  dy2 
PP~  VAx*  -f  Ay2     ^  L      Ay2 
Ax2 


V 


obviously  approaches  1.  ^P',  which  is  equal  numerically  to  Ay  —  dy,  is 
an  infinitesimal  of  higher  order  than  Ax  and  hence  than  the  chord  PP  ; 
hence  QP/PP  approaches  0  and  thus  the  limit  of  the  right-hand  mem- 
ber is  1.     Hence  the  middle  term  approaches  1,  q.  e.  d. 


INFINITESIMALS  AND   PIFFF.KENTi ALS  101 

Furthermore : 

,a\  ,      rdO  ,      dr 

(4)  sin*=^.      cos*=s? 

the  tangent  PT  being  drawn  in  the  direction  of  the  increasing 
s.     Beside  these  there  is  the  formula  of  §  2 : 

(5)  tan*=fr 

7.  Rates  and  Velocities.  The  principles  of  velocities  and 
rates  were  treated  in  Chap.  II.  We  are  now  in  a  position  to 
deal  with  a  larger  class  of  problems. 

Example  1.  A  railroad  train  is  running  at  30  miles  an  hour 
along  a  curve  in  the  form  of  a  parabola : 

(A)  y2  =  500x, 

the  axis  of  the  parabola  being  east  and  west  and  the  foot 
being  taken  as  the  unit  of  length.  The  sun  is  just  rising  in 
the  east.  Find  how  fast  the  shadow  of  the  locomotive  is  mov- 
ing along  the  wall  of  the  station,  which  is  north  and  south. 

Since  30  m.  an  h.  is  equivalent  to  44  ft.  a  sec,  the  problem 
is  : 

Given      ^  =  44;     to  find     ft. 
dt  dt 

From  (A)  :         2ydy  =  500 dx,         dx  =  ^.    V 


O 


* 


Substituting  this  value  of  dx  in  (1),  §  6,  we  FlG>  31 

get: 

ds*  =  da?  +  dtf  =  ^  +  dy>,         dy- 


2502  V2502+2/2 

Hence,  dividing  through  by  dt  and  writing  for  ds/  dt  its  value, 
we  get : 

dy_    250x44 

dt      V2502  +  2/2' 
In  particular  :       —  =    —  ft.  a  sec,  or  21.2  m.  an  h. 

L^_u=250    v'2 


102  CALCULUS 

Example  2.     A  man  standing  on  a  wharf  is  drawing  in  the 

painter  of  a  boat  at  the  rate  of  2  ft.  a  sec.    His  hands  are  6  ft. 

above  the  bow  of  the  boat.     How  fast  is  the  boat  moving 

through  the  water  when  there  are  still  10  ft.  of  painter  out  ? 

4  Let  r  be  the  number  of  feet  of  painter 

out  at  any  instant.     Then 

dr 


=  -2. 
dt 


Fig.  32  j?or  dr /  dt  gives  the  rate  at  which  r  is  in- 

creasing with  the  time,  and  since  r  is  decreasing,  the  rate  is 
negative. 

We  wish  to  find  the  rate  at  which  P  is  moving.  Let  s 
denote  the  horizontal  distance  PB  of  P  from  the  wharf.  Then 
ds /dt  gives  this  rate  numerically,  but  algebraically  ds/dt  is 
negative.     We  desire,  then,  the  value  of  —ds/dt. 

Since  s  and  r  are  connected  by  the  relation  : 

we  have  2  s  ds  =  2  r  dr, 

_ds  _  _  r  dr  _       2r 
dt  s  dt     yr2_36 

Hence,  finally :   ■[-—']       =f      2r      1       =24  ft.  a  sec. 

The  student  will  note  that  the  method  of  solution  consists 
in  determining  first  the  velocity  at  an  arbitrary  instant,  and 
then  substituting  the  particular  value  r  =  10  into  the  result 
thus  obtained. 

EXERCISES 

1.  A  lamp-post  is  distant  10  ft.  from  a  street-crossing  and 
60  ft.  from  the  houses  on  the  opposite  side  of  the  street.  A 
man  crosses  the  street,  walking  on  the  crossing  at  the  rate  of 
4  m.  an  h.  in  the  direction  toward  the  lamp-post.  How  fast 
is  his  shadow  moving  along  the  walls  of  the  houses  when  he 


INFINITESIMALS   AND   DIFFERENTIALS  103 

is  two-thirds  of  the  way  over  ?     When  he  is  55  ft.  from  the 
houses?  Ans.  6  m.  an  h.  and  96  m.  an  h.,  respectively. 

2.  A  kite  is  150  ft.  high  and  there  are  250  ft.  of  cord  out.  If 
the  kite  moves  horizontally  at  the  rate  of  4  m.  an  h.  directly 
away  from  the  person  who  is  flying  it,  how  fast  is  the  cord 
being  paid  out  ?  Ans.   3^  m.  an  h. 

3.  A  point  describes  a  circle  with  constant  velocity.  Show 
that  the  velocity  with  which  its  projection  moves  along  a  given 
diameter  is  proportional  to  the  distance  of  the  point  from  this 
diameter. 

4.  A  revolving  light  sends  out  a  bundle  of  rays  that  are 
approximately  parallel,  its  distance  from  the  shore,  which  is  a 
straight  beach,  being  half  a  mile,  and  it  makes  one  revolution 
in  a  minute.  Find  how  fast  the  light  is  travelling  along  the 
beach  when  at  the  distance  of  a  mile  from  the  nearest  point  of 
the  beach.  Ans.   15.7  m.  a  m. 

5.  The  sun  is  just  setting  as  a  base  ball  is  thrown  vertically 
upward  so  that  its  shadow  mounts  to  the  highest  point  of  the 
dome  of  an  observatory.  The  dome  is  50  ft.  in  diameter. 
Find  how  fast  the  shadow  of  the  ball  is  moving  along  the 
dome  one  second  after  it  begins  to  fall,  and  also  how  fast  it  is 
moving  just  after  it  begins  to  fall. 


104 


CALCULUS 


EXERCISES 

Determine    the    maxima    and    minima    of    the    following 
functions : 


1. 

x  log  X. 

2. 

X  COS  X. 

3. 

xe~*. 

4. 

xne~x. 

Ans.    A  minimum  when  x  =  .3679. 


Ans 


•  ii 


5.    x2  log 


A  minimum 
9.    sin  2  x  —  x. 
10.    sin  x  cos3 a;, 
cos  a; 


maximum  when  x  =cot  x,  0  <  x  <  l  n ; 

-£ir<X<0. 

(J^.    e_A;<  cos  (n£  +  y). 
^.   x  —  tan  x. 


11. 


17.    x  4-  tan  #. 


6. 


logo; 

7.  sin  a;  -f  cos  x. 

8.  a;  -f  sin  a?. 
21.    Prove  that 


^iri 


COtiC 
X 


18.    «x. 


tana; 
^v^;  x*. 

Q.   excos2a;. 
sin  x  +  cos  x  I  <  V2. 


19. 


« 


20.    tan  x  —  2  sin  x. 


22.  Show  how  to  draw  the  shortest  possible  line  from  the 
point  (f ,  0)  to  the  curve  y  =  |»?. 

23.  A  rectangular  piece  of  ground  of  given  area  is  to  be  en- 
closed by  a  wall  and  divided  into  three  equal  areas  by  parti- 
tion walls  parallel  to  one  of  the  sides.  What  must  be  the 
dimensions  of  the  rectangle  that  the  length  of  wall  may  be  a 
minimum  ? 

24.  Find  the  most  economical  proportions  for  a  conical  tent. 

25.  A  foot-ball  field  2 a  ft.  long  and  2b  ft.  broad  is  to  be 
surrounded  by  a  running  track  consisting  of  two  straight  sides 
(parallel  to  the  length  of  the  field)  joined  by  semicircular  ends. 
The  track  is  to  be  4c  ft.  long.  Show  how  it  should  be  made 
in  order  that  the  shortest  distance  between  the  track  and  the 
foot-ball  field  may  be  as  great  as  possible. 


INFINITESIMALS   AND   DIFFERENTIALS  105 

26.  The  number  of  ems  (i.e.  the  number  of  sq.  cms.  of  text) 
on  this  page  and  the  breadths  of  the  margins  being  given, 
what  ought  the  length  and  breadth  of  the  page  to  be  that  the 
amount  of  paper  used  may  be  as  small  as  possible  ? 

27.  A  statue  10  ft.  high  stands  on  a  pedestal  that  is  50  ft. 
high.  How  far  ought  a  man  whose  eyes  are  5  ft.  above  the 
ground  to  stand  from  the  pedestal  in  order  that  the  statue  may 
subtend  the  greatest  possible  angle  ? 

28.  A  can-buoy  in  the  form  of  a  double  cone  is  to  be  made 
from  two  equal  circular  iron  plates.  If  the  radius  of  each 
plate  is  a,  find  the  radius  of  the  base  of  the  cone  when  the 
buoy  is  as  large  as  possible.  Ans.   aV-f. 

29.  At  what  point  on  the  line  joining  the  centres  of  two 
spheres  must  a  light  be  placed  to  illuminate  the  largest  pos- 
sible amount  of  spherical  surface  ? 

@>  A  block  of  stone  is  to  be  drawn  along  the  floor  by  a  rope. 
Find  the  angle  which  the  rope  should  make  with  the  horizontal 
in  order  that  the  tension  may  be  as  small  as  possible. 

Ans.    The  angle  of  friction. 

31.  Into  a  full  conical  wine-glass  whose  depth  is  a  and 
generating  angle  a  there  is  carefully  dropped  a  spherical  ball 
of  such  a  size  as  to  cause  the  greatest  overflow.  Show  that  the 
radius  of  the  ball  is 

a  sin  a 
sin  a -f- cos  2  a 

32.  The  illumination  of  a  small  plane  surface  by  a  luminous 
point  is  proportional  to  the  cosine  of  the  angle  between  the 
rays  of  light  and  the  normal  to  the  surface,  and  inversely  pro- 
portional to  the  square  of  the  distance  of  the  luminous  point 
from  the  surface.  At  what  height  on  the  wall  should  a  gas- 
burner  be  placed  in  order  to  light  most  brightly  a  portion  of 
the  floor  a  ft.  distant  from  the  wall  ? 

Ans.   About  -J-^a  ft.  above  the  floor. 


106  CALCULUS 

33.  A  gutter  whose  cross-section  is  an  arc  of  a  circle  is  to  be 
made  by  bending  into  shape  a  strip  of  copper.  If  the  width 
of  the  strip  is  a,  find  the  radius  of  the  cross-section  when  the 
carrying  capacity  of  the  gutter  is  a  maximum.  Ans.  a/ir. 

34.  If,  in  the  preceding  problem,  the  cross-section  of  the 
gutter  is  to  be  a  broken  line  made  up  of  three  pieces  each 
4  in.  long,  the  middle  piece  being  horizontal,  how  wide  should 
the  gutter  be  at  the  top  ?  '  Ans.   8  in. 

35.  A  wall  27  ft.  high  is  64  ft.  from  a  house.  Find  the 
length  of  the  shortest  ladder  that  will  reach  the  house  if 
one  end  rests  on  the  ground  outside  the  wall. 

36.  A  long  strip  of  paper  8  in.  wide  is  cut  off  square  at  one 
end.  A  corner  of  this  end  is  folded  over  onto  the  opposite 
side,  thus  forming  a  triangle.  Find  the  area  of  the  smallest 
triangle  that  can  thus  be  formed. 

37.  In  the  preceding  question,  when  will  the  length  of  the 
crease  be  a  minimum? 

38.  The  captain  of  a  man-of-war  saw,  one  dark  night,  a 
privateersman  crossing  his  path  at  right  angles  and  at  a 
distance  ahead  of  c  miles.  The  privateersman  was  making 
a  miles  an  hour,  while  the  man-of-war  could  make  only  b  miles 
in  the  same  time.  The  captain's  only  hope  was  to  cross  the 
track  of  the  privateersman  at  as  short  a  distance  as  possible 
under  his  stern,  and  to  disable  him  by  one  or  two  well-directed 
shots ;  so  the  ship's  lights  were  put  out  and  her  course  altered 
in   accordance  with  this   plan.      Show   that   the   man-of-war 

crossed  the  privateersman' s  track  -  V<x2  —  b2  miles  astern  of 
the  latter. 

If  a  =  b,  this  result  is  absurd.     Explain. 

39.  Find  the  area  of  the  smallest  triangle  cut  off  from  the 
first  quadrant  by  a  tangent  to  the  ellipse 


x' 


H-^-  =  l.  Ans.  ab. 

b2 


INFINITESIMALS   AND   DIFFERENTIALS  107 

40.  Assuming  that  the  values  of  diamonds  are  proportional, 
other  things  being  equal,  to  the  squares  of  their  weights,  and 
that  a  certain  diamond  which  weighs  one  carat  is  worth  $  m, 
show  that  it  is  safe  to  pay  at  least  $8m  for  two  diamonds 
which  together  weigh  4  carats,  if  they  are  of  the  same  quality 
as  the  one  mentioned. 

41.  A  man  is  out  in  a  power-boat  a  miles  from  the  nearest 
point  A  of  a  straight  beach.  He  wishes  to  reach  a  point  inland 
whose  distance  from  the  nearest  point  B  of  the  beach  is  b  miles. 
The  distance  AB  is  c  miles.  If  he  can  make  v1  m.  an  h.  in  his 
boat,  but  can  walk  only  v2  m.  an  h.,  what  point  of  the  beach 
ought  he  to  head  for  in  order  to  reach  his  destination  in  the 
shortest  possible  time? 

Ans.   - -  =  • -,    where  &i  and  02  are  *ne  angles  that  his 

*i  v2 

paths  make  with  a  normal  to  the  beach. 

This  problem  is  identical  with  that  of  finding  the  path  of  a 
ray  of  light  that  traverses  two  media  separated  by  a  plane  sur- 
face, as  for  example  when  we  look  at  an  object  submerged  in 
water. 

42.  Assuming  the  law  of  the  refraction  of  light  stated  in 
the  last  problem,  show  that  a  ray  of  light  in  passing  through  a 
prism  will  experience  the  maximum  deflection  from  its  orig- 
inal direction  when  the  incident  ray  at  one  face  and  the  re- 
fracted ray  at  the  other  make  equal  angles  with  their  respec- 
tive faces. 

0.  A  steel  girder  25  ft.  long  is  moved  on  rollers  along  a 
passageway  12.8  ft.  wide,  and  into  a  corridor  at  right  angles 
to  the  passageway.  Neglecting  the  horizontal  width  of  the 
girder,  find  how  wide  the  corridor  must  be  in  order  that  the 
girder  may  go  round  the  corner.  Ans.  5.4  ft. 

44.  A  town  A  situated  on  a  straight  river,  and  another  town 
B,  a  miles  further  down  the  river  and  b  miles  back  from  the 
river   are  to  be  supplied  with  water  from  the  river  pumped 


108  CALCULUS 

by  a  single  station.  The  main  from  the  waterworks  to  A 
will  cost  $  m  per  mile  and  the  main  to  B  will  cost  $  n  per  mile. 
Where  on  the  river-bank  ought  the  pumps  to  be  placed? 

45.  When  a  voltaic  battery  of  given  electromotive  force 
(E  volts)  and  given  internal  resistance  (r  ohms)  is  used  to 
send  a  steady  current  through  an  external  circuit  of  R  ohms 
resistance,  an  amount  of  work  (  W)  equivalent  to 

E2R 


(r  +  K? 


X  107  ergs 


is  done  each  second  in  the  outside  circuit.  Show  that,  if  dif- 
ferent values  be  given  to  R,  W  will  be  a  maximum  when 
R  =  r. 

46.  Show  that,  if  a  point  describe  a  curve  y  =  f(x)  with  a 
constant  or  variable  velocity  v,  the  rates  at  which  its  projec- 
tions on  the  coordinate  axes  are  moving  will  be  respectively : 

dx  dy 

—  =  v  cos  t,  -£  =  v  sin  t. 

dt  dt 

If  the  velocities  of  its  projections  along  the  axes,  namely 
dx/dt  and  dy/dt,  are  known,  then 


*-JK  +  *,  tanr  =  ^/^. 

dt       *dt2      dt2'  dt  I  dt 


47.  If  in  the  preceding  problem  the  curve  be  given  in  polar 
coordinates,  r=f(0),  and  we  consider  two  lines  parallel  and 
perpendicular  respectively  to  the  radius  vector  at  any  instant, 
the  rates  at  which  its  projections  on  these  lines  are  moving 
will  be : 

dr  ,  d6  .     , 

—  =  v  cos  \b,  r  —  =  v  sin  \b. 

dt  Y  dt  r 

48.  A  projectile,  moving  under  the  force  of  gravity,  de- 
scribes a  parabola : 

x  =  v0  cos  a  •  t,         y  =  v0  sin  a  •  t  —  16 12, 


INFINITESIMALS  AND   DIFFERENTIALS 


109 


provided  that  the  resistance  of  the  atmosphere  has  no  appre- 
ciable influence  on  the  motion.  Here,  a  denotes  the  initial 
angle  of  elevation  and  v0  the  initial  velocity.  Determine  the 
velocity  of  the  projectile  in  its  path. 

Ans.    W—  64  v0  sina  •  t  + 1024 12 


49.  A  ladder  25  ft.  long  rests  against  a  house.  A  man 
takes  hold  of  the  lower  end  and  walks  away,  carrying  it  with 
him,  at  the  rate  of  2  m.  an  h.  How  fast  is  the  upper  end 
descending  when  the  man  is  8  ft.  from  the  house  ? 

50.  A  conical  filtering  glass  is  nearly  filled  with  water. 
The  water  is  running  out  of  an  opening  in  the  vertex 
at  a  constant  rate.  How  fast  is  the  surface  of  the  water 
falling  ? 

51.  Let  AB,  Fig.  33,  represent  the  rod  that  connects  the 
piston   of   a   stationary  engine  with   the 
fly-wheel.     If  u  denotes  the  velocity  of  A 
in  its  rectilinear  path,  and  v  that  of  B  in 
its  circular  path,  show  that 

u  =  (sin  6  -f  cos  6  tan  <f>)  v. 


Fig.  33 


52.  Find  the  velocity  of  the  piston  of  a  locomotive  when 
the  speed  of  the  axle  of  the  drivers  is  given. 

53.  A  draw-bridge  30  ft.  long  is  being  slowly  raised  by 
chains  passing  over  a  windlass  and  being  drawn  in  at  the  rate 
of  8  ft.  a  minute.  A  distant 
electric  light  sends  out  hori- 
zontal rays  and  the  bridge 
thus  casts  a  shadow  on  a 
vertical  wall,  consisting  of 
the  other  half  of  the  bridge, 
which  has  been  already  raised. 
Find  how  fast   the    shadow 

is  creeping  up  the  wall  when  half  the  chain  has  been  drawn 
in. 


Fig.  34 


110  CALCULUS 

54.  The  sun  is  just  setting  in  the  west  as  a  horse  is  running 
around  an  elliptical  track  at  the  rate  of  m  miles  an  hour.  The 
axis  of  the  ellipse  lies  in  the  meridian.  Find  the  rate  at 
which  the  horse's  shadow  moves  on  a  fence  beyond  the  track 
and  parallel  to  the  axis. 

55.  Differentiate  y  when 

2x  sin  y  =  3y  sin  x.         Ans.     gg  =  3ycosa»-2siny. 

dx     2x  cosy  —  3  sin  x 

56.  Differentiate  y  when 

y  =  x\og(x-y). 

57.  Plot  the  curve 

r  =  acos30, 

determining  where  the  tangent  is  parallel  to  the  axis  of  a  lobe. 

58.  Plot  the  following  curves  : 

(a)  y  =  x-\-sinx.  (c)  r  =  asin30. 

(b)  y  =  xe~\  (d)     r  =  i. 

d 

59.  Locate  the  roots  of  the  equation 

x  =  cot  x 

and  hence  discuss  completely  the  maxima  and  minima  of  the 
function  in  Question  2. 

60.  The  equation 

0(l  +  cos0)=2sin0 

has  one  root  in  the  interval   —  ir/2  <  6  <  v/2,  namely  0  =  0. 
Has  it  others  ? 

61.  Find  all  the  roots  of  the  equation 

(l+&2)tan-16  =  6. 


CHAPTER  VI 


INTEGRATION 


1.  The  Area  under  a  Curve.  Let  it  be  required  to  compute 
the  area  bounded  by  the  curve 

(i)  y=m, 

the  axis  of  x,  and  two  ordinates  whose  abscissas  are  x  =  a 
and  x  =  b,  (a<  b).  We  can  proceed  as  follows.  Consider  first 
the  variable  area,  A,  bounded  by  the  first  three  lines  just 
mentioned  and  an  ordinate  whose  abscissa  x  is  variable.  Then 
A  is  a  function  of  x.  For,  when  we  assign  to  x  any  value 
between  the  limits  a  and  b  in  question,  the  corresponding 
value  of  the  area  is  thereby  determined  and  could  actually  be 
computed  by  plotting  the  figure  on  squared  paper  and  counting 
the  squares,  or  by  cutting  the  figure  out  of  a  sheet  of  paper 
or  tin  and  weighing  the  piece. 

If,  then,  we  can  obtain  an  analytic  expression  for  this 
function  of  x,  holding  for  all  values  of  x  from  a  to  b,  we  can 
then  set  x  =  b  in  this  formula  and  thus  solve  the  above 
problem. 

To  do  this,  begin  by  giving  to  x  an  arbitrary  value,  x  =  x0, 
and  denoting  the  corre-  y 
sponding  value  of  A  by  A0. 
Next,  give  to  x  an  incre- 
ment, Ax,  and  denote  the 
corresponding  increment  in 
A  by  A  A.  We  can  ap- 
proximate to  the  area  AA 
by  means  of  two  rectangles,  as  shown  in  the  figure,  and  thus 
we  get: 

y0  Ax  <  A^l  <  (y0  +  Ay)  Ax. 
Ill 


Fig.  35 


112  CALCULUS 

Hence  y0  <  — -<  y0  +  Ay. 

Ax 

(Iif(x)  decreases  as  x  increases,  the  inequality  signs  will  be 
reversed.)  Allowing  now  Ax  to  approach  0  as  its  limit,  we  see 
that  the  variable  AA/Ax  always  lies  between  the  fixed  quan- 
tity y0  and  the  variable  y0  +  Ay,  whose  limit  is  yQ.     Hence 

,.      AA 

Ax=0   AX 

or,  dropping  the  subscripts,  we  have : 

(2)  DxA  =  y. 

For  example,  let  the  curve  be 

y  =  x* 
and  let  a  =  1,  b  =  4.     Then 

DxA  =  x2 

and  the  question  is  :  What  function  must  we  differentiate  in 
order  to  get  x2?  We  readily  see  that  Xs /3  is  such  a  function. 
But  this  is  not  the  only  one.  For,  if  we  add  any  constant, 
a?/ 3  +  C  will  also  differentiate  into  x2.  We  shall  see  later 
that  this  is  the  most  general  function  whose  derivative  is  x2, 
and  hence  A  must  be  of  the  form  : 

(3)  A=i+C- 

This  formula  is  not  wholly  definite,  for  C  may  be  any  con- 
stant. On  the  other  hand  we  have  not  as  yet  brought  all  our 
data  into  play,  for  we  have  as  yet  said  nothing  about  the  fact 
that  the  left-hand  ordinate  shall  correspond  to  the  abscissa  x  =  l. 
Now  the  variable  area  A  will  be  small  when  x  is  only  a  little 
greater  than  1,  and  it  will  approach  0  as  its  limit  when  x  ap- 
proaches 1.  If,  then,  (3)  is  to  be  a  true  formula,  it  must  give 
0  as  the  value  of  A  when  x  =  1,  or 

(4)  .0  =  i  +  C,  0=-J. 


INTEGRATION  113 

(5)  ,:  ^=f-|. 

Having  thus  found  the  variable  area,  we  can  now  obtain  the 
area  we  set  out  to  compute  by  putting  x  =  4  in  (5) : 

The  process  of  finding  the  area  under  a  curve  is  thus  seen 
to  be  as  follows.  First  find  a  function  which  when  differen- 
tiated will  give  the  ordinate  y=f(x)  of  the  curve  (1)  before 
us  ;  and  add  an  undetermined  constant  to  this  function.  Next, 
determine  this  constant  by  requiring  that  A  shall  =  0  when 
x  =  a.  Thus  the  variable  area  is  completely  expressed  as  a 
function  of  x.     Lastly,  substitute  x  —  b  in  this  formula. 

EXERCISES 

1.  Show  that,  if  the  area  in  the  foregoing  example  had  been 
measured  from  the  ordinate  x  =  2,  the  value  of  the  constant  C 
would  have  been  —  21 : 

Xs 
A  =  --2%: 
3        3' 

and  if  it  had  been  measured  from  the  origin,  then  C  would 
have  been  =  0 : 

2.  If,  in  (1),  y=f(x)  =x,  the  curve  is  a  straight  line;  and 
if  a  =  6,  b  =  20,  the  figure  is  a  trapezoid.  Compute  its  area 
by  the  above  method  and  check  your  result  by  elementary 
geometry. 

3.  Find  the  area  under  the  curve 

V  =  A 

lying  between  the  ordinates  whose  abscissas  are  a  =  10  and 
a;  =  20.  Ans.  620,000. 


114  CALCULUS 

4.  Find  the  area  of  one  arch  of  the  curve 

y  =  sin  x.  Ans.  2 

5.  Find  the  area  under  that  portion  of  the  curve 

2/  =  l  — JB2 
which  lies  above  the  axis  of  x. 

6.  A  river  bends  around  a  meadow,  making  a  curve  that  is 
approximately  a  parabola : 

y  —  x  —  4sc2, 

referred  to  a  straight  road  that  crosses  the  river,  as  axis  of  x ; 
the  mile  is  taken  as  the  unit.  How  many  acres  of  meadow 
are  there  between  the  road  and  the  river?        Ans.   6  J,  nearly. 

2.  The  Integral.  In  the  preceding  chapters  we  have  treated 
the  problem :  Given  a  function ;  to  find  its  derivative.  The 
examples  of  the  last  paragraph  are  typical  for  the  inverse 
problem:  Given  the  derivative  of  a  function;  to  find  the 
function.     Stated  in  equations,  the  problem  is  this.     If 

Dx  U=  u,  or  dXJ—u  cfcc, 

where  u  is  given,  to  find  U. 

The  function  U  is  called  the  integral  of  u  with  respect  to  x 
and  is  denoted  as  follows  : 

U=  I  u  dx. 

Thus  we  have  the  following 

Definition  op  an  Integral.  The  function  U  is  said  to  be 
the  integral  of  u : 


JJ=  I  udxy 


if  DxU=u,  or  dU=udx. 

The  given  function  u  is  called  the  integrand. 


INTEGRATION  115 

More  precisely,  we  should  say  that  U  is  an  integral  of  u\ 
for  U-\-  G  is  evidently  an  integral,  too,  C  being  any  constant. 
For  example  : 

(6)  Cx*dx  =  -?^-  +  Ci  n*-l. 

J  n  +  1 

For,  if  we  differentiate  the  function: 


/y.M+1 

U=^—  +c 
n  +  1 

with  respect  to  x,  we  get : 

DxU=xn, 

and  xn  is  the  integrand  u  of  the  integral  in  question. 

The   following   theorem   is   fundamental  in  the  theory  of 
integration. 

Theorem  A.   If  two  functions  have  the  same  derivative : 

(A)  Dxf(x)  =  Dx<Mx), 
they  differ  from  each  other  only  by  a  constant. 

Since  the  derivative  of  their  difference  is  0 : 

the  theorem  is  equivalent  to  the  following :  If  the  derivative 
of  a  function  is  always  0 : 

(B)  DMv)=0, 

the  function  is  a  constant. 

Geometrically,  the  truth  of  this  last  theorem  is  exceedingly 
plausible.     The  graph  of  the  function 

is  a  straight  line  parallel  to  the  axis  of  x,  and  its  slope  is  0. 
If,  now,  conversely,  the  slope  of  a  curve  is  always  0,  what  can 
the  curve  be  other  than  a  straight  line  parallel  to  the  axis  of 
x  ?  For  an  analytic  proof  of  these  theorems  cf .  the  chapter 
on  the  Law  of  the  Mean. 


116  CALCULUS 

From  Theorem  A  it  follows  that  all  the  integrals  of  a  given 
function  differ  from  one  another  only  by  additive  constants. 
For,  if  £7  and  U'  are  any  two  integrals  of /(#) : 

D.U=m,  DxU'=f(x), 

then  U  and  U'  have  the  same  derivative. 

Differentiation  and  integration  are  inverse  processes,  and  so 
we  have : 

Dxjudx  =  u  or  d  I  u  dx  =  u  dx; 

JDxUdx  =  U+  G  or  fd(J=  U+  C. 

Theorem  I.  A  constant  factor  can  always  be  taken  out  from 
under  the  sign  of  integration : 

(I)  /  cudx  —  c  I  u  dx. 

Consider  the  two  functions  that  enter  in  (I).    We  have : 

Dx  I  cu  dx  =  cu, 

Dx\  c  /  u  dx   =  c  Dx  I  udx  =  cu, 

i.e.,  the  derivatives  of  these  functions  are  identical.     Hence 
the  functions  themselves  can  differ  at  most  by  a  constant: 


/  cu  dx  =  c  I  udx+k, 


and  so  by  choosing  the  constant  of  integration  in  the  integral 
on  the  left-hand  side  suitably,  we  can  make  k  =  0. 

Theorem  II.     The  integral  of  the  sum  of  two  functions  is 
equal  to  the  sum  of  their  integrals  : 

(II)  I  (u-\-v)dx=  I  udx-\-  I  v  dx. 


INTEGRATION  117 

The  proof  is  lire  that  of  Theorem  I : 

Dz  I  (w  -f  v)  dx  =  u  -f  v, 

Da    J  udx+  I  vdx\  =  Dx  f  udx  +  Dx  I  vdx  —  u-\-  v. 

Hence  the  functions  on  the  two  sides  of  (II)  differ  at  most  by 
a  constant,  Jc.  The  constants  in  any  two  of  the  integrals  may 
be  chosen  at  pleasure  and  then  the  constant  in  the  third  inte- 
gral can  be  so  taken  that  /  •  =  0. 

Integration  of  Polynomials.    By  the  aid  of  the  above  theorems 
any  polynomial  can  be  integrated.     For  example : 

I  (a  +  hx  -f  ex2)  dx—'  I  adx+  I  bxdx+  I  ex2  dx 

=  a  I  dx  +b  I  xdx+c  f  x2  dx 

x2  Xs 

=     ax     -f     o-~  +C-+C. 
Z  o 

» 

Area  under  a  Curve.    We  can  now  express  the  area  discussed 
in  §  1  in  the  form : 


(?) 


A—  fydx  or  A=  jf(x)dx. 


EXERCISES 

Evaluate  the  following  integrals. 

1.  ii3-±x-$x*)#x.  Am.  3x-2xz~-x9+C. 

2.  /  sfxdx.  Ans.   faj*  +  (7. 


dx. 


118  CALCULUS 

e.  m?*  *  8.  ri+*+**,, 

J     Vx  J  X 

>  9i   Find  the  area  above  the  positive  a,xis  of  x  bounded  by 
the  curve : 

10.   Find  the  area  enclosed  between  the  two  parabolas : 
y  =  x2,  y*=x. 

3.  Special  Formulas  of  Integration.  Corresponding  to  the 
Special  Formulas  of  Differentiation,  of  Chap.  IV,  we  can  write 
down  a  list  of  special  formulas  of  integration,  by  means  o"f 
which,  together  with  the  general  methods  discussed  in  this 
chapter,  all  the  simpler  integrals  can  be  evaluated.  Each 
formula  can  be  proven  by  differentiating  each  side  of  the 
equation. 

Special  Formulas  of  Integration 


/■ 


x- 


n+l 


1.  I  xndx  =  -,  n^P—l. 

n  +  l 


I- 


smxdx  =   —cos  a;. 

3.  I  cos  xdx  =  sin  x. 

/dx        . 
--log*- 

5.  iedx  —  ex. 

6.  T-^-o  =  tan"1* 
Jl  +  x* 


INTEGRATION  119 


dx 


■Vl-x2 


=  sin_1a;, 


=  —  cos_1a; 


sec2xdx  =  tana. 
9.  I  osc2xdx  =  —cot a;. 


/• 


To  these  may  be  added  the  formulas  : 

10.  C      dx        =  vers"1  a;. 
J  ^2x-x> 

11.  Ca'dx  =  r-^-- 
J  log  a 

We  have  omitted  the  constant  of  integration  each  time  for 
the  sake  of  simplicity.  But  the  student  must  not  forget  to 
insert  it  in  applying  these  formulas  in  a  given  example. 
Moreover,  we  have  not  included  the  formula: 


/• 


Odx=C. 


4.  Integration  by  Substitution.  Many  integrals  can  be  ob- 
tained from  the  special  formulas  of  §  3  by  introducing  a  new- 
variable  of  integration.  The  following  examples  will  illus- 
trate the  method. 

Example  1.     To  find       I  V«  +  bx  dx. 

Let  a  +  bx  =  y.        Then     b  dx  —  dy 

and  -y/a  +  bx  dx  —  -y^  dy^^\s 

Integrating  each  side  of  this  equation,  we  get : 

C^a~+bxdx=:j-  fy*dy  =  *y*+C. 


120  CALCULUS 

Hence  fvG+ta  dx  =  2V<"  +  te)'  +  G. 

Example  2.     To  find       I  cos  ax  dx. 
Let  ax  =  y.        Then     adx  =  dy, 

cos  axdx  =  -  cos  ?/  cfa/, 
a 

and      lcosaxdx  =  -  I  cosy  dy  =  - sin  y-f  (7=  -sin ax+  C. 
J  aj  a  a 

Example  3.     To  find       /  x  ^/tf  +  x2  etc. 
Let  x2  —  y.         Then    2xdx  =  dy, 

x-VaFTx^  dx  =  xVtf^f^=\^aT+~y  dy, 

and  I  x^/a2  +  x2dx  =  ±  I  Va2  +  y  dy. 

This  last  integral  is  a  special  case  of  the  integral  of  Ex- 
ample 1.  For,  if  the  a  of  that  formula  is  replaced  by  a2,  the 
b  by  1,  and  the  x  by  y,  we  have  the  present  integral.     Hence 

jx  VaFTx2  dx  =  \{a2  +  x2)  *  +  G. 

We  might  have  set        a2-{-x2  =  y. 

Example  4.     To  find        I  tan  x  dx. 
Here     funxdx  =   /**"»*«   f  ~  **  cos  *  =  -  log  cos  s  +  C. 

J  J       COS  X  J         COS£ 

In  substance,  we  have  introduced  a  new  variable,  cos  x  =  #. 
But  in.  practice  it  is  often  simpler,  as  here,  to  refrain  from 
actually  writing  a  new  letter. 


INTEGRATION  121 

In  the  above  examples  we  have  tacitly  assumed  that  if  x 
and  y  are  functions  one  of  the  other,  and  if  f(x)  and  <j)(y)  are 
two  functions  such  that 

f(x)dx  =  <f>(y)dy, 
then  jf(x)  dx  =  j  <j>  (y)  dy. 

We  can  justify  this  assumption  without  difficulty.     For 
Dxff(x)dx=f(x), 

A JV  (y)  dy  =  Dyj<j>  (y)  dy  -Dxy  =  <j>  (y)  Dxy ; 

and  since  /(#)  =  <£  (y)  -%- , 

ctx 

it  follows  that  the  above  integrals  differ  from  each  other  at 
most  by  a  constant,  k.  Hence,  if  the  constant  of  integration 
in  one  of  these  integrals  is  chosen  at  pleasure,  the  constant  of 
integration  in  the  other  can  be  so  determined  that  k  =  0. 

This  theorem  in  integration  corresponds  to  Theorem  V  of 
Chap.  II  in  differentiation.  And,  as  in  the  case  of  that 
theorem,  the  use  of  differentials,  —  and  it  is  to  this  fact  that 
their  importance  is  due,  —  reduces  the  theorem  in  form  to  an 
algebraic  identity : 

/•*-/[*S]j 

EXERCISES 

Evaluate  the  following  integrals  : 

1.  /  Vl  -  x  dx.  Ans.     _  |(1  _«>)*•+ C. 

2.  l^/l  +  2xdx.  Ans.     f(l  +  2aj)*+C. 


122  CALCULUS 


4-     /   4/  *      6.     I  sin  ax  dx.  8.     /  sin  (ttx  +  v)  dx. 

J  Va  +  bx  J  J  7J 

J  a2 


-far*  a  a 

==  •  J.ws.     sin-1  - 

x2  a 


10.      C      dx       .  ^n«.     sin-^+O. 

J  Va2 


ii.  r  gda?  .  ^rcs.  _v^=^+o. 

12.      I  x2y/a3  +  xidx.    14.     / %--         16.     I  a;  sin  x2  dx. 

J  J  a  +  bx  J 

/'      ,  _                .  _       (*  xdx           ,  _,      f*    dx 
xexdx.  15.      I         .     •        17.      /- -• 

18.     I  cot  #  cte.  J.«s.     log  sin  x  -f-  O. 

5.  Integration  by  Ingenious  Devices. 

Example  1.    To  find    I  cos2 Odd. 
Set  cos20  =  i  (1  rhjcojL2  6). 

Then       /cos20  cW  =  }  f(l  +  cos  2  ff)  dO  =  1 0+  J  sin  2  0  +  C, 
(8)  .-.    j cos20 dO  =  ±(0  +  smO cos 0)  + C. 

We  can  now  evaluate  an  important  integral,  namely : 

,  /  Va2  —  x2  dx. 

Let  x  =  a  sin  0 ;  c?sc  =  a  cos  0  d$, 


^a2-x2dx  =  a2cos2$d6} 


INTEGRATION  123 

/  Va2-x2dx  =  a2 J  cos20  d0=^($  +  sin  0  ens  0)  +  C, 
(9)         .-.   /  Vo2"^2"  tte  =  i  r*VS^a?+  a2  sin"1^  J+  G 


Example  2.    To  find   T-/a 


The  integrand  can  be  written  in  the  form : 

_j_=j_r^ L.T. 

a2  —x2     2a  \_x  -f-  a     x  —  aj 

Hence         f-i*-.  =  1-  CJSL.  _  J_  fj?L 

Jas  —  x2     2aJ  x-\-a     2aJ  x  —  a 

=  ~riog(aJ  +  a)-log(«-a)l+C7; 

Example  3.   To  find   f-^ 
J  s 


First  Method. 


J  sin  6     J  , 


sini 

c?0 


0        # 

2  sin  -  cos  - 

2       2 


*In  case  —  a<x<a,  formula  (10)  involves  the  logarithm  of  a  negative 
quantity.  We  can  avoid  this  difficulty  by  writing  the  second  term  in  the 
bracket  as  +  l/(a  —  as),  the  corresponding  integral  thus  becoming 

f_*L= -tog  («-*). 

J  a  —x 

This  leads  to  the  formula : 

(10')  f    dx     =.liog«±g-K7. 

It  will  be  shown  later  in  the  Calculus  that,  in  the  domain  of  imaginaries, 
the  logarithms  of  (10)  and  (10')  differ  from  each  other  only  by  an  imagi- 
nary constant,  and  since  the  latter  may  be  included  in  the  constant  of 
integration,  (10)  and  (10')  may  be  regarded  as  equivalent  formulas. 


!24  CALCULUS 

6 
Let  -  =<£.     Th      the  last  integral  b< 

d<t>       _   J*  sec2  <f>d$__   /Vtaj 

C  dQ       .      .      # 
/  x— 7;=  log  tan- +( 

Second  Method. 


f # fsec2<t>d<l>       Anan<£     i      .      J     _ 

J  sin  *  cos  S     J      tan<r"~e/la^=1°gtan^  +  a 

(11)        ;.:■■;     .;./s^=logtan|+a 
J  sin  (9     J  " 


sin  fl  c?6> 

sin20 

■   r^cos^    =_llogl  +  cos^ 
Jl-cos20         2     gl-^o7^+C# 


The  fraction 


20 
l  +  cosfl_CQS  2         1 

i-cos0~    •  2e~7~70 
sm2^     tan2- 


(11)  .'.    f-^=logtan^+G 


sin  0 


EXERCISES 


1.  Jsm26d0.  Ans.   i(0-smOcosO)  +  G. 

2.    C    de    .  3    /"_*_ 

J  1  +  cos  0_\^  '    J  1  -  COS  0 

4'/cS'  ^logtaag  +  g+C, 


5. 


or    llog^  +  Sm^+fi     or     log  (sec  0-f  tan  0)+  C. 
-L      sin  u 

r   dx  

J  V¥+tf'  log(^  +  V^+a2)-ha 

Suggestion  :  Let  x  =  a  tan  0. 
6*  J  V^=2'  ^   log^-fV^^H-a 


INTEGRATION  125 

6.  Integration  by  Parts.     The  formula  of  d   "erentiation  : 
d  (uv)  =  udv-\-v  du, 
leads  to  a  formula  of  integration : 

(III)  I  udv  =  uv  —  I  v  du. 

Integration  by  means  of  this  formula  is  known  as  integration 
by  parts. 


Example  1.     To  find      I  xexdx. 


Let  u  =  x,  dv  =exdx; 

then  du  =  dx, 

and 


v=  I  exdxz=ex, 
I  xexdx  =  xex  —   I  exdx  =  (x  —  1)  ex  +  C. 

/  / 

Example  2.     To  find       I  log  x  dx. 

Let  u  =  log  x,  dv  —  dx\ 

then  du  = — ,  v=x, 

x 

and  j  log x  dx  =  x  log a;  —  /#  —  =  #(log#  —  1)+  C. 

Example  3.     To  find       /  Va2  +  x?dx. 


Let  w  =  Va2  +  x2,  dv  =  dx; 

xdx 


then  du  =      xdx     ,  v  =  x 


fVair+tfdx  =  x^/a2  +  x2-  f-^ 
J  J  Va2 


4-x2 
Again,  Va2  4-  x*  =  —  , 

VV  +  iC2 


126  CALCULUS 

■dx 


fVtf  +  ^dx  =  a2  C-**—±  f-£ 

J  J  -Va2  +  x?     J  Va2 


+  X2 

Adding  these  two  equations  and  recalling  Ex.  5  in  the  preced- 
ing Exercises,  we  have : 


(12) 


/  Va2  +  x2dx  =  i[xVa2  +  x2  +  a2log(x  +  ^a2  +  x2)]+C. 


EXERCISES 

Evaluate  the  following  integrals : 


1.  I  xeaxdx.  5.  Ixcosaxdx.  9.  I  x  tan"1  x  dx. 

2.  I  xPe^dx.  6.  I  sin-lxdx.  10.  I  xlogxdx. 

3.  /  x*eaxdx.  7.  /tan-1  x  dx.  11.  I  eaxsmxdx. 

4.  Ixsmxdx.  8.  I  xsin^xdx.  12.  I  eaxcosxdx. 


13.     I^a?  —  a2dx. 


Ans.    %[x  Vx2  —  a2  — a2  log  (#  -f-  V&*2  —  a2)]  +  O. 

7.  Use  of  the  Tables.  The  integrals  that  ordinarily  arise  in 
practice  and  which  can  be  evaluated  in  terms  of  the  elemen- 
tary functions  can  be  found  in  such  a  table  of  integrals  as 
Professor  Peirce's,*  and  for  this  reason  it  is  not  necessary  for 
us  to  go  further  into  the  theory  of  integration  in  this  course. 
We  have  learned  how  to  differentiate  all  the  elementary 
functions,  but  not  all  these  functions  can  be  integrated  in 
terms  of  the  elementary  functions.     Thus  the  integral : 

dx  (*         d(b 

or  ' 


f         d4>  (0<&2<1), 

J  Vl-&2sin2<£ 


*  B.  O.  Peirce,  A  Short  Table  of  Integrals,  Revised  Edition,  1899  or 
later,  Ginn  &  Co.,  Boston. 


INTEGRATION  127 

leads  to  a  new  class  of  transcendental  functions,  the  Elliptic 
Integrals,  and  cannot  be  evaluated  in  terms  of  algebraic 
functions,  sines  and  cosines,  etc. 

There  are,  however,  large  classes  of  functions  that  can  be 
integrated,*  and  the  classes  that  are  important  in  practice 
have  been  tabulated.  The  student  is  requested  to  examine 
with  care  the  classification  in  the  Tables  above  referred  to. 

Example  1.     To  find  by  aid  of  the  Tables    C  ®dx    . 

The  integrand  is  a  rational  function  of  x,  and  so  we  look 
under  "II.  Rational  Algebraic  Functions,"  p.  5.  There  we 
find  "A.  —  Expressions  Involving  a -\-bx."  Formula  31  gives 
us  the  integral  we  want : 


/ 


Example  2.     To  find    /  - 


xdx  1      .         1  .  „ 

(l-xy~~      l-x     2(l-af)2        * 

dx 


+  X  +  X2 


Here  the  integrand  involves  rationally  an  expression  of  the 
form  X  =  a  -f  bx  -f  ex2,  and  so  we  look  under  C,  p.  10.  Two 
formulas,  67  and  68,  give  this  integral.  But  since  #  =  4ac 
—  b2  =  3  is  positive,  the  second  formula  would  introduce  imagi- 
naries.     The  first  gives  : 


/: 


^_  =  _l_tan-?£±l+a 
1  +  x  +  x2     V3  V3 


Example  3.     To  find    /  - 
J  ■ 


VI  +  x  +  x2 

Here  the  integrand  involves  VX,  and  so  we  look  under 
"  III.  Irrational  Algebraic  Functions,"  and  find  under  D,  p.  23, 
Formulas  160,  161.     Since  c  =  1  >  0,  we  choose  No.  160 : 

*  When  we  say,  a  function  can  be  integrated,  we  mean,  can  be  inte- 
grated in  terms  of  the  elementary  functions.  Every  continuous  function 
has  an  integral,  for  the  area  under  its  graph  is  an  integral. 


128  CALCULUS 


/ 


dx 


log/Vl+a  +  ^  +  a  +  ^Wo. 


VI  +  x  +  x2  \ 

Example  4.     To  find    /  sin6 a;  dx. 

The  integrand  is  a  transcendental  function.     Turning  to  V, 
p.  35,  and  looking  down  the  list  we  come  to  No.  263  : 

/\  •   n     ,            sinn-1#  cos  x  ,  n  —  1   {*  •   n*     j 
smnx  dx  = ■  -\ I  smn  2x  dx. 
n                  n    J 

If  we  set  here  n  =  6,  we  reduce  the  given  integral  to  an .  ex- 
pression involving    l  sin4sc  dx,  and  this  integral  can  in  turn  be 

reduced  by  the  same  formula,  written  for  n  =  4.     Thus  we  get 
finally  : 


/■ 


sin6#  dx 


sin5 a;  cos  cc     5 sin3 a;  cos  x     5sin#cos  x     5x      ^ 
6~  ~W  "16  16  "* 


Example  5.     To  find    f- — dx 


4  cos  x 
Formula  300  gives : 

r^^  =  |tan-[3tan|]+a 
J  5—  4  cos  x     3  2  J 

EXERCISES 

Evaluate  the  following  integrals  with  the  aid  of  the  Tables. 

+  a 


Ans-  h 


log  (4  —  5x)  + 


/xdx 
(4,-Bxf 

,    r   ax 

"  .Mi-*) 

/dx  .       f*      xdx 

(ar-2)(a>-3)"  '   Ja-2-5z  +  6 


5x 


Ans. h  log h  C. 

x  1  —  x 


INTEGRATION  129 


j5  +  3^  V15  1^5;^ 

/*      da? 

J  »  +  ^  + 


*» 


^na.     i  log ^ — L  tan"1  ?  ^—  +  0. 

9.     f^-=^dx.         Ans.     2Vr^  +  logA/l~a;~1-K7. 
«/        *  VT^oJ  +  1 

10.  r-^=.  12.  /v^e?* 

11.  f— p=r  13.  rvf+^^ 

t/  a;VlH-ar  t/  a; 

14.      /  V—  1-f  4a  —  a2  (to. 

\2  7  2  V3 

15.  r * — r       is.  r    d*     . 

J  (7  -9X  +  2X2)*  J  xVtf+px  +  q 

17.     f dx  18.     fSm2Ocos2$dO. 

J(l-^)Vl  +  ^  J 

8.  Length  of  the  Arc  of  a  Curve.  We  have  seen  in  Chap.  V, 
§  6,  that  the  differential  of  the  arc  of  a  curve  is  given  by  the 
formula : 

ds=  Veto2  +  dy2  =  \/l  +^dx. 
*        dx2 

Hence  the  length  of  the  arc  can  be  obtained  by  integration, 
and  we  have : 

<13>  s=f\k+%dx- 


130  V  CALCULUS 

Example  1.     Let  us   find  the   length   of  the    arc    of    the 
parabola : 

y  =  x\ 


Here        dy  —  2x  dx,  -Vdx2  -f-  dy2  =  Vl  4-  4 x2  dx, 

s  =   I  Vl  +  4:X2  dx, 

and  this  integral  can  be  reduced  at  once  to  (12)  in  §  6,  or  to 
Formula  124  of  the  Tables: 

s  =  2J  Vi  +  x2 dx  =WiT?  +  ilog(>+  y/±+¥)+  a 

If  we  measure  the  arc  from  the  vertex,  s  =  0  when  x  =  0,  and 
we  have  for  the  determination  of  C : 

0  =  ±log4+C,  '(7=1  log  2. 


Hence         s  =  \x  V1  +  4&2  +  \  log  (2x  +  V 1  +  4a;2). 

In  particular,  the  length  of  the  arc  to  the  point  (1,  1)  is 

M*=i  =  iV5  +  ilog(2-fV5). 

On  p.  Ill  of  the  Tables  we  find  a  table  of  natural  logarithms, 
from  which  we  see  that 

log  (2  +  V5)  =  log  4.24  =  1.45. 

Hence  [s]x=1  =  1.48. 

As  a  check  on  this  result  we  note  that  the  length  of  the 
chord  is  V2  =  1.41;  on  the  other  hand,  the  length  of  the 
broken  line  consisting  of  the  abscissa  and  the  ordinate  of 
the  point  (1,  1)  is  2.  Consequently  the  length  of  the  arc  in 
question  must  lie  between  1.41  and  2. 

Example  2.   To  find  the  length  of  the  arc  of  the  equiangular 

spiral : 

r  =  aeke<  X  =  cot  a. 


Here  ds  =  Vdr2  +  rW,         dr  =  aXe^dO, 


INTEGRATION  131 


ds  =  VT+X2ae^  dO  =        ,+  X  dr  =  dr  sec  a, 

A 


=  sec  a  I  dr  = 


r  sec  a  +  k. 


If  we  measure  the  arc  from  the  point  0  =  0,  r  =  a,  then  s  =  0 
when  r  =  a  and 

0  =  a  sec  «  4-  k.         .-.         s  =  (r  —  a)  sec  a. 

When  0  =  —  cc ,  the  spiral  coils  round  the  pole  r  =  0  infinitely 
often,  and  r  approaches  0  as  its  limit.  The  value  of  s,  taken 
numerically,  when  r  <  a,  is : 

|  s  |  =  — s  =  (a  —  r)  sec  a. 

Thus  we  see  that  the  length  of  the  spiral  does  not  increase 
beyond  all  limit  when  0  =  —  go,  but 

lim   |  s  |  =  a  sec  a. 

0=-co 

EXERCISES 

1.  Find  the  length  of  the  cardioid  : 

r  =  a (1  —  cos  6).  Ans.  8 a. 

2.  Find  the  length  of  the  spiral  r  =  0  from  the  pole  to  the 
point  where  it  crosses  the  prime  vector  for  the  first  time, 
0  =  2tt.  Ans.  21.3. 

3.  Find  the  length  of  the  arc  of  the  curve  27  y2  =  a?  included 
between  the  origin  and  the  point  whose  abscissa  is  15. 

Ans.  19. 

"4.  Find  the  length  of  the  arc  of  the  spiral  r  =  1/6,  measured 
from  the  point  0  =  1,  r  =  1. 

5.   Prove  that  the  length  of  the  arc  of  the  catenary  -J 
measured  from  the  vertex,  x  =  0,  is  :     s=  -lea—  e  aJ- 


132  CALCULUS 

6.   Assuming  the  equation  of  a  parabola,   referred   to   the 
focus  as  pole,  in  the  form : 

m 
1  —  cos<£' 

find  the  perimeter  of  the  segment  cut  off  by  the  latus  rectum. 
Check  your  answer. 

EXERCISES 

Obtain  the  following  integrals  without  the  aid  of  the  Tables. 

sin  x  dx 


+  b  cos  x 


dx 

ex  +  e~x ' 

'sin  OdO 
cos2  6 


1.     fV2^xdx.           9.     A**"**.  17.     C-™ 

J                                    J        x  N  J  a  + 

r  dx                 /v  4-1  r 

'    2.     I                            10.     /  — HJi(tt.  18.     I  ecosx  sin  a;  do;. 

3.  f(a-xfdx.        11.     /*— —  •  19.     Csm3xdx. 

4.  frcfi-xtydx.  *12.     C^p^ldx.  20.     f- 

5-  JrS-      i3-  J1™*-  2l-  /- 

J  l  +  a>                        J  v             ;  Jl  +  sma 

7.  T^icte.           15.     C^~~1^dr.  '23.     fsetfxdx. 

8.  I 16.     I  xcosafdx.  24.     I  cos3xdx. 

J  a  +  bx*                    J  J 

25.   Let  J[  denote  the  area  bounded  by  the  curve 

a  fixed  radius  vector  6  =  00,  and  a  variable  radius  vector  0  =  $, 
see  Fig.  29.     Show  that 

and  thus  obtain  the  theorem : 


INTEGRATION  133 


(14) 


i  Cr*dO. 


26.  Find  the  area  of  the  cardioid : 

r  =  a  (1  —  cos  <f>) .  Ans.   f  ira?. 

27.  Determine  the  area  cut  out  of  the  first  quadrant  by  the 
arc  of  the  equiangular  spiral  r  =  aeKe  corresponding  to  values 
of  6  between  0  and  %tt. 

28.  Obtain  the  area  of  one  lobe  of  the  lemniscate : 

r2  =  a2  cos  2  0. 

29.  The  same  for  r  =  a  sin  36. 

30.  The  same  for  r  =  a  cos  nO. 

31.  Find  the  area  of  the  ellipse : 

—  +  ^  =  1.  Ans.   nab. 

a2     b2 

32.  Prove  that  the  length  of  the  arc  of  the  curve 

a2 

a2  — or 

measured  from  the  origin,  is 

t      a-\-x 

s  =  a  log  — ■ x. 

a  —  x 

33.  Prove  that  the  length  of  the  arc  of  the  curve 

Sa2y2=x2(a2-x2)       is       s  =  y  +  —  sin"1  - . 
'34.   Prove  that  the  area  of  the  curve 

[y J  =a2  —  x2      is      ira2. 

J  35.   Determine  the  area  of  the  loop  of  the  curve 

y2  =  x2  +  x3'  Ans.  T87 


CHAPTER  VTT 
CURVATURE.    EVOLUTES 

1.  Curvature.  We  speak  of  a  sharp  curve  on  a  railroad  and 
thus  express  a  qualitative  characteristic  of  the  curve.  Let  us 
see  if  we  cannot  get  a  quantitative  determination  of  the  degree 
of  sharpness  or  flatness  of  curves  in  general. 

If  we  consider  the  angle  <£  by  which  the  tangent  of  a  curve 

has  changed  direction  when  a  point  that  traces  out  the  curve 

has  moved  from  P  to  P,  then  this  angle  will 

depend,  not   only   on   the   sharpness   of   the 

curve,  but  also  on  the  distance  from  P  to  P'. 

We  can  nearly  eliminate  this  latter  element 

Fig.  86  when   P'  is   near  P  by  taking  the   average 

change  of  angle  per  unit  of  arc,  <j>/PP\     This  ratio  we  define 

as  the  average  curvature : 

-£-  =  average  curvature  for  arc  PP. 
PP' 

The  limit  approached  by  this  average  curvature  is  what  we 
understand  by  the  curvature  at  P;  it  is  denoted  by  k  : 

(1)  k  =  lim  -^-  —  actual  curvature  at  P. 


p'=p 


PP 


Thus  for  a  circle  of  radius  a, 

PP'=od>,         -£--.,        lim-i^l 
PP>      (i         p'=p  ppi     a 

134 


CURVATURE.    EVOLUTES  135 

and  the  average  curvature  does  not  change  with  P.  The 
curvature  of  a  circle  is  the  same  at  all  points  and  is  equal  to 
the  reciprocal  of  the  radius.  Again,  the  curvature  of  a  straight 
line  is  0. 

To  evaluate  the  limit  (1)  for  any  curve,  y  =f(x),  we  observe 
that,  if  we  write 

PP'=AS,  <f>  =  ±T, 

then  k  =  lim  —  =  Dsr, 

a«=o  As 

where  t  denotes  as  usual  the  angle  which  the  tangent  of  thB 
curve  makes  with  the  axis  of  x.  More  precisely,  it  is  the 
numerical  value  of  Dar  which  we  want,  for  k  is  an  essentially 
positive  quantity  (or  0).     Hence 

(2)  k=±—  ,    or  better:     K  =  \  — 

w  ds  Ids 

From  the  foregoing  definition  we  see  that  the  curvature  is 
the  rate  at  which  the  tangent  turns  when  a  point  describes 
the  curve  wifh  unit  velocity. 

To  compute  dr/ds  we  have 

(3)  tanr  =  ^       or       r  =  tan"1^. 

dx  dx 

It  will  be  convenient  to  introduce  a  shorter  notation  for  deriva- 
tives and  we  shall  adopt  Lagrange's,  which  employs  accents : 

y  *=/(?), 

It  follows,  then,  that 

dy'  =  (&dx  =  (^)dx  =  y"dx 
dx  dx2 


and  ds  =  Vdx2  -f  dy2  =  VI  -\-y'2  dx. 

Returning  to  (3)  and  differentiating  we  have: 


136 


CALCULUS 


r-taa-V,        ^=«^  =  fJE 


dr 
ds 


i+y'2    i+y'2) 


.'/ 


(l+y,2P 


(4) 


I  y"  I 


(1+2/'2) 


[ 


♦*r 


The  reciprocal  of  the  curvature  is  called  the  radius  of  curva- 
ture and  is  usually  denoted  by  p :  * 

l__(l+y")f    L      dtf 

\y"\    " 


(?) 


/>  =  K  = 


dx2 


;  The  radius  of  curvature  of  a  circle  is  its  radius.  The  curva- 
ture of  a  curve  at  a  point  of  inflection  is  in  general  0 ;  for 
y"  =  0  at  such  a  point  if  y"  is  continuous  there. 

Example.     To  find  the  curvature  of  the  parabola 


f 


2mx. 


Here 


2ydy  =  2mdx,  y' 


y 


dy'  =  --2dy, 

y 


__  ™?\y 


m" 


=  (m2  +  y*)? 


(m2  + 


*  The  student  can  always  recall  which  of  these  two  ratios  is  the  curva- 
ture, which  the  radius  of  curvature,  by  the  check  of  dimensions.  If  we  re- 
gard x  and  y  each  as  of  the  first  degree  in  length,  then  y'  =  dy/dx  is  of  the 
0-th  and  y"  —  dy' /dx  of  the  —  1st  degree.  Hence  the  bracket  is  of  the  0-th 
degree  and  |  y"  |  of  the  —  1st,  and  the  ratio  must  therefore  be  written  so 
as  to  yield  p  of  the  1st,  k  of  the  —  1st  degree  in  length. 


CURVATURE.    EVOLUTES  137 

EXERCISES 
Find  the  curvature  of  each  of  the  following  curves. 


1. 

1   2. 
L  3. 

A. 

5. 
^6. 

y=x*. 

y=3?,  at  the  origin. 
y  =  log  esc  x. 

The  ellipse:  2?  +  £«.l. 
a2     6^ 

The  hyperbola:  —  —  ^-  =  1. 
a2     b2 

The  equilateral  hyperbola : 

xy 

Arts,     k  = • 

(1  +  4^)2 

Ans.     K  =  |sinic]. 

4—                    «4&4 

(6V  +  aV)f 
a2 

2* 

(y  +  2/2)2 

7.  Show  that  the  radius  of  curvature  of  the  curve  y  —  x^ 
approaches  0  as  its  limit  when  the  point  P  approaches  the 
cusp,  (0,  0). 

8.  Find  the  radius  of  curvature  of  the  curve 

54  y  =  10a5  -  19x4  +  11jb8+  x2  -  12x 
at  the  origin.  Ans.     p  =  125. 

9.  Find  the  radius  of  curvature  of  the  catenary 

at  the  vertex.  Ans.     p  =  a. 

10.  At  what  points  of  the  curve  y=x?  is  the  curvature 
greatest  ? 

11.  Find  the  locus  of  the  points  in  which  the  curvature  of 
the  curves  of  Fig.  4  :  y  =xn,  x  >  0,  n>  0,  is  greatest. 


138  CALCULUS 

2.  The  Osculating  Circle.  At  an  arbitrary  point  P  of  a  curve 
let  the  normal  be  drawn  toward  the  concave  side  of  the  curve 
and  let  a  distance  be  laid  off  on  this  normal  equal  to  the  radius 
of  curvature,  p.  The  point  Q  thus  obtained  is  called  the  centre 
of  curvature.  The  circle  constructed  with  Q  as  centre  and  with 
radius  p  stands  in  an  important  relation  to  the  curve.  It  is 
called  the  osculating  circle  and  has  the  property  that  it  repre- 
sents the  curve  more  accurately  near  P  than  any  other  circle 
does.  Consider  the  family  of  circles  drawn  tangent  to  the 
curve  at  P  and  with  their  centres  on  the  concave  side.  Those 
whose  radii  are  very  short  are  curved  too  sharply,  —  more 
sharply  than  the  given  curve.  Now  let  the  circles  grow.  If 
we  pass  to  the  other  extreme  of  circles  with  very  large  radii, 
these  will  be  too  flat.  Evidently,  then,  certain  intermediate 
circles  come  nearer  to  the  shape  of  the  curve  at  P  than  these 
extreme  ones  do.  It  is  not  difficult  to  find  a  criterion  by 
means  of  which  one  circle  is  characterized  as  better  than  all 
the  others.  Draw  the  tangent  at  P  and  drop  a  perpendicu- 
lar from  P'  on  it  meeting  it  in  M  and  cutting  an  arbitrary 
one  of  the  circles  in  R.  Then,  as  we  shall  show 
later,  MP  will  in  general  be  an  infinitesimal 
of  the  second  order  referred  to  the  arc  PP  as 
principal  infinitesimal,  and  PR  will  also  be  of 
the  second  order  for  a  circle  taken  at  random. 

We  can,  however,  in  general  find  one  circle  for 
Fig.  37 

which  PR  will  be  an  infinitesimal  of  the  third 

order,  and  it  turns  out  that  this  circle  is  precisely  the  oscu- 
lating circle.  We  shall  give  the  proof  later  (Chap.  XIII,  §  9). 
The  osculating  circle  cuts  the  curve  in  general  at  the  point 
of  tangency ;  but  there  may  be  certain  exceptional  points  at 
which  this  is  not  the  case.  Near  such  a  point  P'R  is  an  infini- 
tesimal of  even  higher  order  than  the  third,  in  general,  of  the 

fourth. 

EXERCISE 

Construct  carefully   the   parabola  y  =  x2  for   values  of  x : 
—  |  <  x  <  f ,  taking  10  cm.  as  the  unit.     Draw  the  osculating 


CURVATURE.     EVOLUTES 


139 


circle  at  the  point  x  =  £,  y  =  \,  and  also  at  the  vertex.  Ink  in 
the  parabola  in  a  fine  black  line,  the  first  osculating  circle  in 
red,  and  the  second  in  a  different  colored  ink  or  in  pencil. 

3.  The  Evolute.  When  a  point  P  traces  out  a  curve,  the 
centre  of  curvature  Q  traces  out  a  second  curve.  This  latter 
curve  —  the  locus  of  Q — is  called  the  evolute  of  the  given 
curve.  We  proceed  to  deduce  its  equation  and  to  discuss  its 
properties. 

The  point  Q  can  be  found  analytically  by  writing  down  the 
equation  of  the  normal  at  P  and  determining  the  intersection 
of  this  line  with  a  circle  of  radius  p,  hav- 
ing its  centre  at  P.  The  equation  of  the 
normal  is 

(6)  X-x  +  y'(Y-y)  =  0, 

where  (X,  Y)  are  the  running  coordinates, 

i.e.  the  coordinates  of  a  variable  point  on 

the  normal,  and  (a?,  y)  the  coordinates  of 

P,  —  the  latter  being  held  fast  during  the  FlG*  3b 

following  investigation.     The  equation  of  the  circle  is 


(?) 


(x-xy+(Y-yy=P* 


a+y,z) 


To  find  where  (6)  and  (7)  intersect,  eliminate  X : 


(l+y>*)(Y-yy 


a  +  y'2) 
y"2    ' 


Y-y 


i+y2 

y" 


Which  sign  must  we  take?     Notice  that  when  the  curve  is 
concave  upward,  as  in  the  figure, 


Y-y>0 


and 


a  _  <i2y 


dx2 


Hence  in  this  case  we  must  use  the  upper  sign : 


140  CALCULUS 

On  the  other  hand,  when  the  curve  is  concave  downward, 
Y-y<0        and        2/"=g<0, 

and  again  we  have  the  upper  sign.  Hence  (8)  is  always  true 
and 

y 

From  (6)  and  (8)  we  get : 

s=,-y'(i+y"). 

y 

The  values  of  X  and  Fthus  found  are  the  coordinates  (x1}  y^ 
of  the  point  Q,  and  so  we  have : 

dx\        dxr)  dxr 

(9)      «.=* — v~-  »-»+-35- 

do2  dx2 

These  formulas  involve  no  radicals. 

If  we  eliminate  x  and  y  between  the  two  equations  (9)  and 
the  equation  y  =f(x)  of  the  given  curve,  we  shall  obtain  the 
equation  of  the  evolute  in  the  form 

F(x1,yl)  =  0. 

But  it  is  not  necessary  to  eliminate.  We  can  plot  as  many 
points  on  the  evolute  as  we  like  by  substituting  in  (9)  the 
values  of  x,  y,  y',  and  y"  corresponding  to  successive  points  on 
the  given  curve. 

Example  1.     To  find  the  evolute  of  the  parabola 
(10)  y2  =  2mx. 

Here        ^/  =  ™         1  +  df  =  m^±f         ^l==_m2. 
dx      y'  dx2  y2  dx2  y3 

Hence  tt1  =  a;-m^  +  ^)  /-^  =  x  +  ^±l, 

y3         I        y3  m 


CURVATURE.    EVOLUTES 


141 


y2     I        ys  m2 

and  it  remains  to  eliminate  x  and  y  between  these  equations 
and  (10).     Eliminating  x  we  have  : 


m 

2m 
3 

From  the  second  equation : 


y2    ,  ra2-h  y2  .  Sy2 

2m  m  2m 


or 


(xi-mj^y2. 


m2yl=f. 


To  eliminate  ?/  between  these  last  two  equations,  square  each 
side  of  the  last  and  cube  each  side  of  the  preceding  one.  Thus 
we  get : 

4    2     8mV  V 

Vl       27~\Xl~mJ' 

Dropping  the  subscripts  we  have  as  the 
equation  of  the  evolute  of  the  parabola  :    q 

8 


01) 


f 


(  x  —  m) . 
27m\  J 

This  is  a  so-called  semi-cubical  parabola.  FlG  3y 

Example  2.     To  find  the  evolute  of  the  ellipse 


(12) 


S+*-l. 

9      •       7  9  * 


We  obtain  without  difficulty  the  equations  : 

dy  _      b2x 
dx         a2y' 


d?y  =        b4 
dx2  a2!/3' 


Xl~X  tfb2         ; 


2/i  =  2/ 


y(Va* +  <*?)_ 


a2b4 


To  eliminate  x  and  y  between  these  equations  and  (12)  requires 
a  little  ingenuity.     From  (12)  we  have 


142 


CALCULUS 


tfx2  +  a2y2  =  a2b2,  a4y2  =  a?b2  (a2  —  x2), 

b'x2  +  «y  =  62(tt4  -  aV  +  b2x2). 

Hence  s,  =  ,  -  ^"^  +  ^  =  <^  * 

a462  a4 

In  a  similar  manner  we  get : 

a2y(b4-b2y2  +  a2y2)  _       a2 


3/i  =  2/ 


a2/>4 


—  2Z3. 


We  can  solve  these  equations  respectively  for  x2  and  y2  and 

substitute   the   values    thus    ob- 
tained in  (12) : 


/    ax,   \t      /    byY    \ 
\a2—b2)   ^{tf-b2) 


=  1. 


Fig.  40 


Dropping  the  accents  we  have  as 
the  final  equation  of  the  e  volute 
of  the  ellipse : 

(13)     (ooj)*  +  (6y)*  =  (a2  -  62)l 


EXERCISES 

Find  the  equation  of  the  evolute  of  each  of  the  following 
curves. 

x2     ?/2 


1.   The  hyperbola: 


=  1. 


Ans.     (as)*  -  (byy  =  (a2  +  b2)  K 

2.  The  hyperbola  :  2xy  —  a2. 

Ans.     (x  +  y)%  —  (x—  y)%  =  2c$. 

3.  The  catenary  :       y  =  ^(ex  +  e~x). 

Ans.     xx  =  x  —  \  (e2x  —  e~2x),     yl  =  2y  ; 

2\4 


-*(i±v5 


4.    ic^  +  y*  =  a*. 


Ans.     (x  +  #) $  +  (a;  —  y)»  =  2cA 


CURVATURE.     EVOLUTES 


143 


Ans.     x2  +  ?/2 


5.  a;  =  a  (cos  0  +  #  sin#), 
y  =  a(sinO  —  $  cos0). 

6.  x  =  acos30,     y  =  asin36.     Ans.    (x  +  y)1* -\- (x  —  y)* 


4.  Properties  of  the  Evolute.  The  property  of  the  evolute 
to  which  the  curve  owes  its  name  is  the  following.  Suppose  a 
material  cylinder  to  be  constructed  on  the  concave  side  of  the 
evolute  and  a  string  to  be  wound  on  the  cylinder,  Fig.  41. 
Let  a  pencil  be  fastened  to  the  end  of  the  string,  the  point 
being  placed  at  a  point  P0  of  the  given  curve  and  the  string 
drawn  taut  and  fastened  at  a  point  A  of  the  evolute  so  that  it 
cannot  slip.  If  now  the  pencil  is  moved  along  the  paper  so 
that  the  string  unwinds  from  the  evolute  or  winds  up,  the 
pencil  will  describe  the  given  curve. 

To  prove  this,  let  P  be  an  arbitrary  point  of  the  given 
curve,  Q  the  corresponding  point  of  the 
evolute,  and  P'  the  position  of  the  pencil  \g^r 
when  the  string  leaves  the  evolute  at  Q. 
We  wish  to  prove  that  P'  coincides  with 
P.  To  do  this  it  is  sufficient  to  show 
(a)  that  QP  is  tangent  to  the  evolute,  so 
that  P'  lies  on  QP;  and  (b)  that  QP'  = 
QP=P. 

ad  (a)     Writing  equations   (9)  in  the 
form  : 

y  y 

and  differentiating  with  respect  to  x,  we  have  ;  * 


<tei-r,-.y'(i+y'*)y'" 


dx 


f 


Sy'2 


y'0-  +  y'*)y'"-3y»y' 
y"2 


-g  -  yii  =  3y>  _  (1  +  y-)  ^_  -  _, 


r 


// 


*  The  student  may  find  it  more  convenient  in  working  out  these  differ- 
entiations to  retain  the  form  (9).     Lagrange's  form  is  more  compact. 


144 


CALCULUS 

dyl  _      1  _ 
dxx  ~      y'~ 

1 

dyJ 

dx 

and  thus  the  slope  of  the  evolute  at  Q  is  seen  to  be  the  ne?*  ■ 
tive  reciprocal  of  the  slope  of  the  given  curve  at  P.     Hei. 
QP  is  tangent  to  the  evolute,  q.  e.  d. 

ad  (b)  If  we  denote  by  s±  the  length  of  the  arc  Q0  Q  of  the 
evolute,  then  (  =  s1  +  p0,  and  we  wish  to  show  that  this 
quantity  is  eqr      to  p  : 

«i  +  po  =  p- 

It  is  evidently  sufficient  to  show  that 

d9i  _  dp 
dx      dx 

Now  dsf  =  dxx2  -f  dyi\ 

+2/if2=^2(i+j!T)=2/i'2(i+y2). 

And  again : 

ft= ±3yy-(u.f).v"VrT7'=  ±*m+F. 

dx  y"2 

Hence  P=±£, 

dx  dx 

and  since  we  have  taken  sx  so  that  it  increases  when  p  in- 
creases, the  upper  sign  holds  : 

ds1  =  dp,  s±  =  p+  C. 

At  Qo,     s1  =  0  and  p  =  p0,     hence  0  =  p0  +  Cy 

and 

P  =  s1  +  Po,  q.e.  d. 

We  have  shown  incidentally  that  the  normals  to  the  given 
curve  are  tangent  to  the  evolute.  Thus  it  appears  that  the 
evolute  is  the  envelope  of  the  normals  of  the  given  curve.  This 
property  can  be  used  as  the  definition  of  the  evolute  and  its 
equation  is  then  readily  deduced  by  the  method  of  Chap.  XVII. 


CURVATURE.    EVOLUTES  145 

EXERCISES 

1.   If  the  equation  of  the  curve  is  given  in  po  ar  coordinates, 
r=/(0),  then  (see  Fig.  29) 

At  =  A^  +  A0 

,  ,  dr      d\b  .  d& 

and  hence  —  =  —  +  —  • 

ds      ds      ds 

Remembering  that 

d$      r 
tan^  =  r—  =  -, 

dr     r1 
where  r'  =  dr/dd,  obtain  the  formula, 

drfl* 


[_        dO2} 


(14)  P=±^     ^r9^ 

d02^    d02 

Find  the  radius  of  curvature  of  each  of  the  following  curves 
at  any  point. 

(r2  +  a2)* 

2.  The  spiral  of  Archimedes  r  =  ad.         Arts,   p  =  v  _      .  \  • 

r  r2  +  2a2 

I. — 

3.  The  cardioid  r  =  2  a  (1  —  cos  <£).  Ans.   p  =  f  V ar. 

a2 

4.  The  lemniscate  7^  =  a2  cos  2  0.  ^ws.   p  =  — . 

or 

r3 

5.  The  equilateral  hyperbola  r2  cos  2  0  =  a2.         ^Lns.   f>  =  -j- 

6.  The  equiangular  spiral  r  =  aeK0. 

7 .  The  trisectrix  r  =  2  a  cos  0  —  a. 

.  a(5-4cos0)* 

S  ^   '  =     9-6cos0    ' 


CHAPTER   VIII 


THE    CYCLOID 


1.  The  Equations  of  the  Cycloid.  The  cycloid  is  the  path 
traced  out  by  a  point  in  the  riui  of  a  wheel  as  it  rolls,  i.e.  by  a 
point  in  the  circumference  of  a  circle  which  rolls  without  slip- 
ping on  a  straight  line,  always  remaining  in  the  same  plane. 
Let  the  given  line  be  taken  as  the  axis  of  x  and  let  6  be  the 
angle  through  which  the  circle  has  turned  since  the  point  P 
was  last  in  contact  with  the  line  at  0.     The  coordinates  of  P, 


Fig.  42 


x  =  OM  and  y  —  MP,  can  be  expressed  as  follows  in  terms  of 
6.  We  notice  that  the  arc  NP=aO  of  the  circle  and  the  seg- 
ment ON  of  the  line  are  of  equal  length,  since  the  circle  rolls 
without  slipping.     Hence 

OM=  ON- 


Also, 

and  so  we  have: 

(1) 


MP=NS 

{x  =a(6 
y  =  a(l 

as  the  equations  of  the  cycloid. 

146 


MN^aO  —  asmS. 
LS  =  a  —  a  cos  0 ; 


sin  0), 
cos  0), 


THE  CYCLOID  147 

It  is  possible  to  eliminate  6  between  these  equations  and 
thus  obtain  a  single  equation  between  x  and  y.  But  the  func- 
tions thus  introduced  are  less  simple  than  those  of  equations 
(1)  and  it  is  more  convenient  to  discuss  the  properties  of  the 
curve  directly  by  means  of  these  equations. 

EXERCISES 

1.  The  equations  of  the  cycloid  referred  to  parallel  axes 
with  the  new  origin  at  the  vertex,  i.e.  the  highest  point,  are : 

J  x  =  ad  +  a  sin  0, 
^  '  \y=—a  +  a cos  0, 

the  angle  6  now  being  the  angle  through  which  the  circle  has 
turned  since  the  point  P  was  at  the  vertex.  Obtain  these 
equations  geometrically,  drawing  first  the  requisite  figure, 
and  verify  the  result  analytically  by  transforming  the  equa- 
tions (1)  : 

x  =  x'  +  -n-a,         y  =  y'-\-2a,         6  =  0'  -f  ir. 


2.    Show  that  the  equations  of  an  inverted  cycloid  referred 
to  the  vertex  as  origin  can  be  written  in  the  form : 

x  =  aO  -f  a  sin  6, 

a  — a  cos  6.  x 


Draw  the  figure  and  interpret  $  geometrically. 


2.  Properties  of  the  Cycloid.     The  slope  of  the  curve  is 

dy  _       a  sin  Odd  sinfl     _  2  sin  \B  cos  \0  _     ,  x  ~ 

dx     adO  —  acosOdO     1  —  cos0  2sm2±0 

or  tanr  =  cot^0. 

From  this, result  we  infer  that  the  tangent  at  P  is  perpen- 
dicular to  the  chord  PN,  Fig.  43.  For  the  latter  makes  an 
angle  of  \6  with  the  negative  axis  of  x  and  hence  its  slope  is 
—  tani0,  i.e.  the  negative  reciprocal  of  the  slope  of  the  tan- 
gent.    Thus  we  see  that  the  normal  at  P  goes  through  the 


148 


CALCULUS 


7 

V 

p)i 

^/2 

S 

/ 

» 

0 

f 

fcsSN. 

Fig.  43 


and 


(6) 


aV 


d 


dy 


dx 
dx 


lowest  point  of  the  generating  circie 
and  hence  the  tangent  at  P  goes 
through  the  highest  point. 

The  equation  of  the  tangent  at 
the  point  (x0,  y0),  $  =  00,  is 

(4)  y-yo  =  w>HOo(F-Xo), 
and  of  the  normal : 

(5)  x-x0  +  cot  \BQ(y  -  2/0)  =  0. 

The  Evolute.     We  have  seen  that 

^=coU0.  Hence  l+^-2=csc2i-0 
dx  dx2  l 


±  csc2  ±0d0 


add  —  a  cos  OdQ 
4asin4i0 


4asin4i0' 


sin3i0 


4a  sin  \B. 


It  is  now  easy  to  construct  the  centre  of  curvature  and  thus 
find  the  evolute.  We  have  merely  to  lay  off  on  the  normal  PN 
a  distance  PQ  =  4asin|0,  i.e. 
double  the  distance  PN.  The 
locus  of  the  point  Q  is  thus  seen 
to  be  an  equal  cycloid  having  its 
vertex  at  the  origin  O.  We  leave 
the  proof,  which  is  simple,  to  the 
student,  referring  him  to  Fig.  43. 

Fig.  44 

The  Arc.     We  have  : 

ds2  =  dx2  +  dy2  =  a2[(l  -  cos  0)2  +  sin20]d02  =  4a2sin2i0d02, 

s  =  2a  /sini0d0  =  4a  / sini0d(i0)  =  -4acos|-0  +  C. 


THE   CYCLOID 


149 


If  we  measure  the  arc  from  the  origin, 

0=-4a  +  C,  C=4a, 

(7)  .-.     s  =  4<x(l-cos£<9)  =  8asin2i0. 

The  total  length  of  one  arch  of  the  cycloid  is,  therefore,  8  a. 

Area  of  an  Arch.  This  area  was  first  determined  experimen- 
tally by  Galileo,  who  cut  out  an  arch  and  weighed  it.  We  can 
find  the  area  under  the  curve  by  integration  : 

A—  I  ydx  =  I  [a  — a  cos  0]  [a  dO  —  acos  0a"0] 
=  a2  r(l-2cos<9  +  cos20)a-0 


(8) 


a2  [0  -  2  sin  $  +  J  (0  +  sin  0  cos  0)]  +  (7, 

0  =  0+<7, 
\     A  =  a2(f  0  -  2sin  0  +  i-  sin  0  cos  0). 


The  area  of  the  complete  arch  is,  therefore,  3ira2,  or  three 
times  that  of  the  generating  circle. 


3.  The  Epicycloid  and  the 
Hypocycloid.  When  a  circle 
rolls  without  slipping  on  a 
second  circle  that  is  fixed, 
always  remaining  in  the 
plane  of  the  latter,  a  point 
P  in  the  circumference  of 
the  moving  circle  traces  out 
an  epicycloid.  From  Fig.  45 
it  is  clear  that 


Fig.  45 


x=  OK+  KM=  (a  +  b)  cos0  +  b  sin[<f>  -(Z-o\\, 
2/=iT^-i>S'  =  (a  +  6)sin0-6cosr^-/'|-0^1. 


150 


CALCULUS 


Furthermore,  the  arc  AN=  a0  and  the  arc  NP=b<f>  are  equal 
a0  =  b<j>.     Hence  we  have  as  the  equations  of  the  epicycloid : 


x  =  (a  +  b)  cos  0  -  b  cos  ^±^0, 


(9) 


y  =  (a  +  b)  sin 0—b sin 


b 

a  +  b, 


If  the  variable  circle  rolls  on  the  inside  of  the  fixed  circle, 
the  path  of  the  point  P  is  a  hypocycloid.  Its  equations  are 
obtained  in  a  similar  manner  and  are: 


(10) 


x  =  (a  —  b)  cos  6  -\-b  cos 6, 

y  =  (a  —  b)  sin  0  —  &  sin 6. 


The  following  special  cases  are  of  interest. 
(1)  If  a  =  26,  the  hypocycloid  reduces  to  a  segment  of  a 
straight  line,  namely,  the  diameter  of  the  circle,  y  =  0.  Thus 
a  journal  on  the  rim  of  a  toothed  wheel  which  meshes  inter- 
nally with  another  wheel  of  twice  the  diameter  describes  a 
right  line,  so  that  circular  motion  is  thereby  converted  into 
rectilinear  motion. 

(2)  When  a  =  4&,  the  equations  of  the 
hypocycloid  reduce  to  the  following  (cf. 
Tables,  Formulas  580  and  585) : 

r       x  =  3b  cosO +bcos30  =  acos30, 
\       y  =  36  sin  6  —  b  sin  3 0  =  a  sin3  6. 


Hence 


xs  +  y1 


Fig  46  This   is   the   equation   of   the  four-cusped 

hypocycloid. 
The  cycloids  play  an  important  part  in  Applied  Mechanics, 
in  the  theory  of  the  shape  in  which  the  teeth  of  gears  should 
be  cut. 

For  a  more  extensive  discussion  of  the  subject  of  this  chap- 
ter see  Williamson,  Differential  Calculus,  Chap.  XIX,  Roulettes. 


THE   CYCLOID  151 

EXERCISES 

1.  Show  by  means  of  the  equation  of  the  normal  of  the 
cycloid,  (5),  that  the  normal  goes  through  the  lowest  point  of 
the  generating  circle. 

2.  Obtain  the  equations  of  the  path  of  the  journal  of  the 
driver  of  a  locomotive  and  plot  the  curve. 

3.  Obtain  the  equations  of  the  path  of  a  point  on  the  outer 
edge  of  the  flange  of  a  driver. 

4.  Obtain  the  equations  of  the  path  of  the  pedal  of  a  bicycle. 

5.  Obtain  the  equation  of  the  path  of  an  arbitrary  point  in 
the  wheels  of  a  sidewheel  steamboat. 

The  curves  of  Exs.  2-5  are  called  trochoids. 

6.  Find  the  velocity,  v,  of  the  point  that  generates  a  cycloid. 

Ans.  v  =  2awsin^0  =  2Fsini0,  where  wis  the 
angular  velocity  of  the  wheel  and  V  the  linear 
velocity  of  the  hub.  At  the  vertex  v=  2V,  i.e. 
the  velocity  of  the  highest  point  of  the  wheel  is 
twice  that  of  the  hub. 

7.  Find  the  area  included  between  an  arch  of  the  cycloid 
and  its  evolute.  Ans.   4:ira2. 

8.  Show  that  the  length  of  the  arc  of  an  inverted  cycloid 
(3),  measured  from  the  vertex  is 

s  =  4asin  t. 

9.  Obtain  the  equations  of  the  evolute  of  the  cycloid  ana- 
lytically, by  means  of  equations  (9)  in  Chap.  VII. 

10.  At  what  points  is  the  trochoid 

x=a$  —  b  sin  6,         y  =  a—bcos6,  (b<a) 

steepest  ?  Ans.   When  cos  0  =  -  . 

a 

11.  Find  the  area  under  one  arch  of  the  trochoid  of  ques- 
tion 10.  Ans.   2t:<£ -\-  7rb2. 


152  CALCULUS 

12.  The  epicycloid  for  which  b  —  a  is  a  cardioid : 

r  =  2a (1  —  cos  <f>), 

the  cusp  being  taken  as  the  pole.     Obtain  this   result  from 
equations  (9). 

13.  Obtain  the  result  in  question  12  directly  geometrically. 

14.  Prove  by  elementary  geometry  that  the  hypocycloid  for 
which  b  —  \a  is  a  straight  line. 

15.  Show  that  the  equation  of  the  normal  of  the  hypocycloid 
is: 

(sin  0O  +  sin  £ZL_  $0)(x  —  x0)  =  (cos  0O  —  cos  ^=—  0O)  (y  —  y0). 

0  0 

16.  Prove  that  the  normal  of  the  hypocycloid  passes  through 
the  point  of  contact  of  the  rolling  circle. 

17.  Work  out  questions  15  and  16  for  the  epicycloid. 

18.  Show  that  the  hypocycloid  for  which  b  =  \a  and  that 
for  which  b  =  j  a  are  the  same  curve. 

19.  Show  that  the  length  of  the  four-cusped  hypocycloid  is 
three  times  the  diameter  of  the  fixed  circle. 


20.   Find  the  area  of  the  four-cusped  hypocycloid. 


Ans.    W 


8 

21.  Find  the  area  enclosed  between  one  arch  of  a  four-cusped 
epicycloid  and  the  fixed  circle.  Ans. 

22.  Obtain  the  equations  of  the  epitrochoid. 

23.  Obtain  the  equations  of  the  hypotrochoid. 

24.  How  many  revolutions  does  the  rolling  circle  make 
in  tracing  out  a  cardioid  ?  a  four-cusped  hypocycloid  ?  How 
many  revolutions  does  the  moon  make  in  a  lunar  month  ? 

25.  How  many  cusps  does  an  epicycloid  have  when  a  and 
b  are  commensurable:  a/b=p/q?  What  can  you  say  about 
this  curve  when  a  and  b  are  incommensurable  ? 


CHAPTER   IX 


DEFINITE    INTEGRALS 

1.  A  New  Expression  for  the  Area  under  a  Curve.  In  Chap. 
VI  we  learned  how  to  compute  the  area  A  under  a  continuous 
curve,  y=f(x),  by  integration.     We  found  tnat 


DxA  =  y 
and  hence  finally  : 


A—   lydx+C, 

We  will  now  consider  a  new  method  of  computing  the  same 
area.  Let  the  interval  (a,  b)  of  the  axis  of  x :  a<x<b,be 
divided  into  n  equal  parts  and  let  ordinates  be  erected  at  each 
of  the  points  of  division.  Let  rectangles  be  constructed  on 
these  subintervals  with  altitudes  equal  to  the  ordinate  that 
forms  their  left-hand 
boundary.  Then  it  is 
obvious  that  the  sum 
of  the  areas  of  these 
rectangles  will  be  ap- 
proximately equal  to 
the  area  A  in  question, 
and  will  approach  A  as 
its  limit  when  n  is  al- 
lowed to  increase  without  limit, 
at  the  end  of  this  chapter. 

153 


Fig.  47 


A  formal  proof  will  be  found 


154  CALCULUS 

We  will  next  formulate  analytically  the  above  sum.     The 
area  of  the  first  rectangle  is 

f(a)Ax  or  f(x0)Ax, 

where  Ax  denotes  the  length  of  the  base,  x1  —  x0  =  (b  —  a)/n. 
The  area  of  the  second  rectangle  is  f(x^)  Ax,  and  so  on.  Hence 
the  sum  of  these  areas  is 

(2)  f(x0)  Ax  +/(»!>  A*  +  ...  +/(a>n-i)  A*, 

and  thus,  allowing  n  to  increase  without  limit,  we  obtain  the 
result : 

(3)  A=*  lim  t/(a*,)Aa>+/(aJi)  Aa>  +  •••  +/<X_1)Az]. 

Example.   Let ' 

2/  =/(«)  =  sin  a?, 

and  let  the  interval  (a,  b)  be  the  interval  O^x^tt/2.  Take 
n  =  10.     Then  Ax  =  3.14/20  =  .157,  and  we  have  to  compute 

sin  0°  Ax  +  sin  9°  Ax  -\ +  sin  81°  Ax. 

Here 

sin   0°=   .000  sin  45°=   .707 

sin    9°=    .156  sin  54°=    .809 

sin  18°=    .309  sin  63°=    .891 

sin  27°=    .454  sin  72°=   .951 

sin  36°=   .588  sin  81°=    .988 

1.507  4.346 

and  thus  we  obtain 

5.853  x. 157  =  .92. 

2.   The    Fundamental    Theorem  of    the   Integral  Calculus. 

Equating  the  two  values  of  A  found  in  §  1  to  each  other,  we 
obtain 

(4)  \im[f(x0)Ax+f(x1)Ax+  ...  +/(^_1)Ax"]=|  |/(aj)cteT 

Although  the  formulas  (1)  and  (3)  were  deduced  from  geomet- 
rical considerations,  the  final  result  (4)  is  purely  analytic  in 


DEFINITE  INTEGRALS  155 

its  character.  We  may  liken  the  process  to  that  of  building  a 
masonry  bridge.  First  a  wooden  arch  is  erected.  On  this  are 
placed  the  blocks  of  granite,  and  when  the  structure  is  com- 
pleted the  wooden  arch  is  removed.  The  bridge  is  the  essential 
thing,  the  wood  was  incidental.  And  so  here  the  geometrical 
pictures  are  but  a  means  to  the  end,  which  is  an  analytical 
theorem, — the  theorem  on  which  the  integral  calculus  rests. 
Let  us  state  the  result  in  words. 

Fundamental  Theorem  of  the  Integral  Calculus.  Let 
f(x)  be  a  continuous  function  of  x  throughout  the  interval 
al£x=b.     Divide  this  interval  into  n  equal  parts  by  the  points 

x0  =  a,  x1}  x2,  "',  xn_ly  xn  =  bj  and  form  the  sum : 

f(x0)\x-{-f(x1)Ax-\ \-f(xn_1)Ax. 

If  n  now  be  allowed  to  increase  without  limit,  this  sum  will  ap- 
proach a  limit;  and  this  limit  can  be  found  by  integrating  the 
function  f(x)  and  taking  the  integral  between  the  limits  x  =  a  and 
x  =  b: 

Expressed  as  a  formula,  the  theorem  is  as  follows  : 
lmf/(«0)Aoj+/(ajl)Aaj+  •••  +/(«„_!) AafUl"   (f(x)dx~X\ 

Instead  of  choosing  the  altitudes  of  the  rectangles  in  §  1 
as  the  left-hand  ordinates,  Ave  might  equally  well  have  taken 
the  right-hand  ordinates.   We  should  then  have  in  place  of  (3)  : 

(5)  A  =  \im\f(xl)Ax+f(x2)Ax+  ...  -f/CaQAaTL 
and  hence  in  place  of  (4) : 

(6)  limf/^Ax-f/^A^-h  -•  +/(»„) A«"]  =  r    Cf{x)dx1T. 

In  fact,  we  might  even  choose  intermediate  ordinates  for  these 
altitudes  if  we  wished. 


156  CALCULUS 

Again,  it  is  not  necessary  to  take  the  subintervals  (x0,  xr), 
(#!,  #2),  •••all  equal.  Their  lengths  Ax0,  Aa?j,  •••  are  arbitrary. 
But  in  that  case  the  longest  of  these  must  converge  toward  0 
when  n  increases  indefinitely. 

Finally,  a  definition.  The  limit  of  the  sum  (3)  or  (5)  is 
called  the  definite  integral  of  the  function /(a),  and  is  denoted 

by 


/ 


f(x)dx. 


In  distinction  from  the  definite  integral,  which  is  the  limit  of 
a  sum,  that  which  we  have  hitherto  called  an  integral,  namely 
the  inverse  of  a  derivative,  is  called  an  indefinite  integral. 

The  integral  sign  had  its  origin  in  the  old-fashioned  long  s, 
the  initial  letter  of  summa,  the  integral  being  thus  conceived 
as  a  definite  integral,  the  limit  of  a  sum. 

Such  a  sum  as  the  one  that  enters  in  (2)  or  (5)  is  frequently 
written  in  the  form : 

n— 1  n 

2}  f(xk)  Ax  resp.  ]jP  /  (xk)  Ax. 


EXERCISES 

1.  Write  out  the  sum  (2)  for  the  interval  0<^#^1  when 

^—1+5        and       Ax=1> 

and  compute  its  value.     Determine  the  limit  of  the  sum  (2) 
for  this  function  by  means  of  the  indefinite  integral. 

2.  Taking  as  the  interval  0  <J  x  <^  J  and  letting 

/(*}=— !=,  A*  =  .05, 

VI  —  ar 


compute 


Ax  J2*        Ax 


SVl-^2  SVl-av2 


DEFINITE   INTEGRALS 


157 


Determine  the  limit  of  the  corresponding  sums  when  Ax 
approaches  0,  and  show  by  a  figure  that  this  limit  Ires  between 
the  two  sums  just  computed. 

3.  Volume  of  a  Solid  of  Revolution.  If  a  plane  curve  rotates 
about  an  axis  lying  in  its  plane,  it  generates  the  surface  of  a 
solid  of  revolution.  Let  us  determine  the  volume  of  such  a 
solid,  its  bases  being  planes  perpendicular  to  the  axis. 

Take  the  axis  of  revolution  as  the  axis  of  abscissas  and 
divide  the  portion  of  the  axis  that  lies  between  the  bases  into 
n  equal  parts  by  the  points 
Xq  =  a,  xx ,  •  •  •  #„_! ,  xn  =  b.  Pass 
planes  through  these  points 
of  division  perpendicular  to 
the  axis,  thus  dividing  the 
solid  up  into  slabs.  We  can 
approximate  to  the  volumes 
of  these  slabs  by  means  of 
cylinders  whose  bases  are  the 
successive  cross-sections.     The  volume  of  the  Zc-th  cylinder  is 

and  the  volume  of  the  solid  in  question  is  thus  seen  to  be  the 
limit  of  the  sum  of  the  volumes  of  these  cylinders : 

F=lim   iry02Ax  +  iryx*&x-\ +  7ryn_12Ax  , 


i.e. 

CO 


=  7T  /  y2dxf 


where  yz=<j>(x)  is  the  equation  of  the  generating  curve. 

For  example,  let  it  be  required  to  find  the  volume  of  a 
segment  of  a  sphere.     Here  the  generating  curve  is  a  circle, 

.      at-hy^T*, 

and,  h  denoting  the  altitude  of  the  segment,  the  abscissas  of 
the  bases  are  r  —  h  and  r.     Hence 


158  CALCULUS 


F  =  tt  ffr2-aAdx 


o          x 
irx 

3 


=  |M 


If,  in  particular,  /i——r,  we  have  the  complete  sphere  and 
obtain  the  familiar  result  f  tt?*3. 

EXERCISES 

1.  Show  that  the  volume  of  an  ellipsoid  of  revolution  is 
|7ra62,  where  a  denotes  the  half-length  of  the  axis. 

2.  A  spindle  is  formed  by  the  rotation  of  an  arch  of  the  curve 

y  =  sin  x 
about  its  base.     Find  its  volume.  Ans.   4.93480. 

3.  Show  that  the  volume  of  a  cone  is  ^irr2h,  and  that  the  vol- 
ume of  a  frustum  is 

4.  Show  that  the  volume  of  a  segment  of  a  paraboloid  of 
revolution,  of  arbitrary  altitude,  is  one-half  that  of  the  circum- 
scribing cylinder. 

.5.  A  cycloid  revolves  about  its  base.  Show  that  the  vol- 
ume of  the  solid  generated  is  57r2a3. 

6.  The  four-cusped  hypocycloid 

x%  +  y$  =  a* 

rotates  about  the  axis  of  x.     Find  the  volume  of  the  solid 

generated.  4        32-rra3 

Ans.    _. 

7.  Find  the  volume  of  a  segment  of  the  solid  of  revolution 
generated  by  the  catenary : 

when  it  rotates  about  the  axis  of  x,  the  plane  x  =  0  forming 
one  of  the  bases.  Am    w(£  _  g-»  +  a 


DEFINITE  INTEGRALS 


159 


4.  Other  Volumes.  We  will  begin  with  the  following  ex- 
ample. A  wood-cutter  starts  to  fell  a  tree  4  ft.  in  diameter 
and  cuts  half  way  through.  One  face  of  the  cut  is  horizontal, 
and  the  other  face  is  inclined  to  the  horizontal  at  an  angle  of 
45°.     How  much  of  the  wood  is  lost  in  chips  ? 

Since  the  solid  whose  volume  we  wish  to  compute  is  sym- 
metric, we  may  confine  ourselves  to  the  portion  OABC.  Divide 
the  edge  OA  into  n  equal 
parts  and  pass  planes  through 
these  points  of  division  per- 
pendicular to  OA.  The  solid 
is  thus  divided  into  slabs  that 
are  nearly  prisms  ;  only  the 
face  QRB'Q'  is  not  a  plane. 
Let  us  meet  this  difficulty 
by  constructing  a  right  prism 
on  PQR  as  base  and  with 
PP'  as  altitude.  Then  its 
volume  will  be  a  little  greater 
than  that  of  the  actual  slab. 
The  solid  formed  by  the  n  prisms  thus  constructed  differs  in 
volume  but  slightly  from  the  actual  solid.* 

We  will  next  formulate  analytically  the  volume  of  the 
prisms.  The  base  PQR  is  a  45°  right  triangle.  Let  OP=xk 
and  PQ  =  yk .     Then,  by  the  Pythagorean  Theorem, 

Hence  the  volume  of  this  prism  is 

iyi?Ax  =  ±(4-x*)Ax, 

*  Let  the  student  not  proceed  further  till  this  point  is  perfectly  clear  to 
him.  Let  him  make  a  model  of  the  actual  solid  out  of  cardboard  or  a 
piece  of  wood  and  draw  neatly  the  lines  in  which  the  plane  sections 
through  P  and  P'  perpendicular  to  OA  cut  the  solid.  He  will  then  be 
able  to  visualize  the  auxiliary  prisms  without  difficulty  and  to  perceive 
that  the  sum  of  their  volumes  approaches  as  its  limit  the  volume  to  be 
computed. 


Fig.  49 


160  CALCULUS 

and  the  volume  of  the  solid  we  wish  to  compute  is 

(8)     lim  [i  (4  -  V)  *b  +  i  (4  -  x*)  A*  +  .  • .  +  J  (4  -  xn_12)  Ax]. 

n=oo 

The  problem  before  us  is  thus  reduced  to  that  of  computing 
the  limit  (8).  Now  inspection  of  this  limit  shows  that  it  is  of 
the  same  type  as  the  limit  (3)  of  §  1 ;  in  fact,  the  two  vari- 
ables become  identical  if  we  put 

Hence  the  limit  (8)  can  be  computed  by  integrating  the 
function  J  (4  —  x2)  and  taking  the  integral  between  the  limits 
x  =  a  =  0  and  x  =  b  —  2 : 


/ 


4(4-^)fe  =  2x-J, 


The  total  volume  is  twice  this  amount,  and  thus  it  appears 
that  there  were  5J  cu.  ft.  of  chips  hewn  out. 


EXERCISES 

1.  A  banister  cap  is  bounded  by  two  equal  cylinders  of 
revolution  whose  axes  intersect  at  right  angles  in  the  plane  of 
the  base  of  the  cap.     Find  the  volume  of  the  cap.      Ans.   -fa3. 

2.  A  Rugby  foot-ball  is  16  in.  long,  and  a  plane  section  con- 
taining a  seam  of  the  cover  is  an  ellipse  8  in.  broad.  Find  the 
volume  of  the  ball,  assuming  that  the  leather  is  so  stiff  that 
every  plane  cross-section  is  a  square.  Ans.   341-J  cu.  in. 

3.  Do  the  preceding  problem  on  the  assumption  that  the 
leather  is  so  soft  that  every  plane  cross-section  is  a  circle. 

Ans.    536  cu.  in. 

4.  A  solid  is  generated  by  a  variable  hexagon  which  moves 
so  that  its  plane  is  always  perpendicular  to  a  given  diameter 
of  a  fixed  circle,  the  centre  of  the  hexagon  lying  in  this  diam- 


DEFINITE   INTEGRALS 


161 


eter,  and  its  size  varying  so  that  two  of  its  vertices  always 
lie  on  the  circle.     Find  the  volume  of  the  solid.    Ans.   2V3a3. 

5.  A  conoid  is  a  wedge-shaped 
solid  whose  lateral  surface  is 
generated  by  a  straight  line  which 
moves  so  as  always  to  keep  par- 
allel to  a  fixed  plane  and  to  pass 
through  a  fixed  circle  and  a  fixed 
straight  line;  both  the  line  and 
the  plane  of  the  circle  being  per- 
pendicular to  the  fixed  plane. 
Find  the  volume  of  the  solid. 

Ans.    %ira2h. 


Fig.  50 


6.  Find  the  superficial  area  of  two  of  the  solids  considered 
above. 

7.  Show  that  the  volume  of  an  ellipsoid  whose  semi-axes  are 
of  lengths  a,  b,  c  is  ^-n-abc. 

5.  Fluid  Pressure.  We  will  next  consider  the  problem  of 
finding  the  pressure  of  a  liquid  on  a  vertical  wall.  Let  the 
surface  be  bounded  as  indicated  in  the 
figure  and  let  it  be  divided  into  n  strips 
by  ordinates  that  are  equally  spaced. 
Denote  the  pressure  on  the  ft-th  strip  by 
APk.  Then  we  can  approximate  to  &Pk 
as  follows.  Consider  the  rectangle  cut 
out  of  this  strip  by  a  parallel  to  the 
axis  of  x  through  the  point  (xk,  yk). 
The  pressure  on  this  rectangle  is  less 
than  that  on  the  given  strip ;  but  we  do 
not  yet  know  how  great  it  is.  Still,  if 
we  turn  the  rectangle  through  90°  about 
its  upper  side,  the  ordinate  yk,  we  shall 
obviously  have  decreased  the  pressure 
further.     Now  the  pressure  on  the  rectangle  in  this  new  posi- 


Xk*l 


Fig.  51 


162  CALCULUS 

tion  is  readily  computed.  It  is  precisely  the  weight  of  a  col- 
umn of  the  liquid  standing  on  this  rectangle  as  base.  The 
volume  of  such  a  column  is  (xk-\- c)ykAx,  and  if  we  denote  by 
w  the  weight  of  a  cubic  unit  of  the  liquid,  then  the  weight  of 
the  column  in  question  is 

w(xk  +  c)ykAx. 
This  is  less  than  APk. 

In  like  manner  we  can  find  a  major  approximation  by  con- 
sidering the  rectangle  that  circumscribes  the  given  strip  and 
whose  altitude  is  yk+i,  and  then  turning  it  over  on  its  lower 
base.     The  pressure  on  it  in  its  new  position  is 

and  this  is  larger  than  APk.     We  thus  obtain : 
(9)  w(xk  +  c)ykAx  <  APk  <w(xk+l  +  c)yk+1Ax. 

If  we  write  out  the  relations  (9)  for  k  =  0,  1,  ••• ,  n  —  1 : 
w  (x0  +  c)  yQ  Ax  <  AP0  <  iv  (x1  +  c)  y1  Ax, 
w  (#j  +  c)  yx  Ax  <  AP1  <  w  (x2  -f  c)  y2  Ax, 


and  add  them  together,  we  see  that  the  pressure  P  we  seek  to 
determine  lies  between 

(10)  w(x0  +  c)y0Ax  +  w(x1  +  c)y1Ax  +  •••  +w(xn_1  +  c)ynlAx 
and 

(11)  w(x1  +  c)y1Ax  +  w(x2-\-c)y2Ax^ \-w(xn  +  c)ynAx. 

Finally,  allow  n  to  become  infinite.     Each  of  the  variables 
(10)  and  (11)  approaches  as  its  limit  the  definite  integral 

i 
w  I  (x  +  c)ydx. 

a 

But  the  pressure  P  always  lies  between  these  variables,  and 
hence  it  must  coincide  with  their  common  limit.  Thus  we  see 
that 


DEFINITE  INTEGRALS  163 

(12)  P  =  w  I  (x  +  c)y  dx. 

a 

We  have  deduced  our  result  under  the  assumption  that  the 
ordinates  of  the  bounding  curve  never  decrease  as  x  increases. 
The  formula  is  true,  however,  even  if  this  condition  is  not  ful- 
filled, as  we  shall  show  in  §  6. 

Example  1.  To  find  the  pressure  on  the  end  of  a  tank  that  is 
full  of  water. 

Here  it  is  convenient  to  take  the  axis  of  y  in  the  surface  of 

the  liquid,  so  that  c  =  0.     The  equation  of  the  bounding  curve 

is  _ 

y  =  k, 

and  thus  h 

_wh2k 

2 


r  x2 

P=w  I  xkdx=ivk 


2 


Now  the  area  of  the  rectangle  is  hk,  so  that,  if  we  write  the 
result  in  the  form  , 

P=w  -hk  •  ^, 

it  appears  that  the  total  pressure  is  the  same  as  what  it  would 
be  if  the  rectangle  were  turned  through  90°  about  a  horizontal 
line  through  its  centre  of  gravity  and  lying  in  its  surface,  and 
thus  supported  a  column  of  the  liquid  of  height  -J-/L 

Example  2.   A  water  main  6  ft.  in  diameter  is  half  full  of 
water.     Find  the  pressure  on  the  gate  that  closes  the  main. 
The  pressure  on  half  the  gate  is 

3 


'/' 


x  V  9  —  xr  dx, 


where  w,  the  weight  of  a  cubic  foot  of  water,  is  62J  lbs.     Turn- 
ing to  Peirce's  Tables,  Formula  135,  we  find  * 

1  The  integral  may  be  evaluated  directly  by  introducing  as  a  new  vari- 
abl  i  of  integration,  y  =  9  —  a2. 


164  CALCULUS 


Hence 


fxV9-xidx  =  --L(9-xt)k 


3 


a;V9  -  x>dx  =  -|(9  -  a2)*  |0  =  9, 
and  the  total  pressure  is  2  x  62  J  x  9  =  1120  lbs. 

EXERCISES 

1.  A  vertical  masonry  dam  in  the  form  of  a  trapezoid  is  200 
ft.  long  at  the  surface  of  the  water,  150  ft.  long  at  the  bottom, 
and  is  60  ft.  high.     What  pressure  must  it  withstand  ? 

Ans.  9300  tons. 

2.  A  cross-section  of  a  trough  is  a  parabola  with  vertex 
downward,  the  latus  rectum  lying  in  the  surface  and  being 
4  ft.  long.  Find  the  pressure  on  the  end  of  the  trough  when 
it  is  full  of  water.  Ans.  66  lbs. 

3.  One  end  of  an  unfinished  watermain  4  ft.  in  diameter  is 
closed  by  a  temporary  bulkhead  and  the  water  is  let  in  from 
the  reservoir.  Find  the  pressure  on  the  bulkhead  if  its  centre 
is  40  ft.  below  the  surface  of  the  water  in  the  reservoir. 

Ans.   Nearly  16  tons. 
6.   Duhamel's  Theorem.   Let 

«i  +  «2  +  ...  •  +  an 

be  a  sum  of  positive  infinitesimals  which  approaches  a  limit  when 
n  becomes  infinite;  and  let 

Pi  +  P*  +  -+Pn 
be  a  second  sum  such  that  fik  differs  from  ak  by  an  infinitesimal 
of  higher  order : 


where 


lim^  =  l, 

A.1+, 
% 

h,             Pk  = 

=  ak  -f  **«*, 

:k  is  infinitesimal. 

Then 

Iim[ft  +  ft4-  ••• 

+  ftj  =  lim  [«!  +  «*+  • 

"+<]. 

DEFINITE  INTEGRALS 


165 


In  accordance  with  the  hypothesis  of  the  theorem  we  have 

4-<i«i  +  «2«8+. henan, 

and  we  wish  to  show  that  the  last  line  of  this  equation  ap- 
proaches 0  when  n  =  cc.  Let  r/  be  numerically  the  largest  of 
the  ^8.     Then 

—  V  ^  ei  ^  V> 

—  V  ^  €^  —  Vf 


y^tn^y. 


—  yct1<€lal^r]au 


< 


—  7jan  ^  eM«;l  ^  ry«. 


< 


Hence 

—  (ai  +  «2+ h«w)^S£iai+ hc»«Bg(al+a2+ |-«n)i/- 

But  17  approaches  0  and  ax  +  «2  +  •••  +  «„  remains  finite.     This 
completes  the  proof. 


Application.  As  a  typical  appli- 
cation of  Duhamel's  Theorem  we 
will  give  the  completion  of  the  proof 
of  formula  (12).  Let  y[  be  the 
minimum  ordinate  in  the  A:-th  strip, 
and  let  y'k'  be  the  maximum  ordinate. 
Then  we  have  by  like  reasoning  to 
that  of  §  5  : 


Vu 


Fig.  52 

w  (xk  +  c)y'k  Ax  <  APk  <  w  (xk+l  +  c)  y'k'  Ax, 
and  hence  P  lies  between  the  two  variables 

(13)  w  (xq  +  c)  yl  Ax  +  w  (xx  +  c)  y[  Ax  -\ \-w  (ajB_i  +  c)yi_i  Aa? 

and 

(14)  !«(#!  + c)^'Aa?  +  w(a2  +  c)2/i'A£H r-w(»»4-c)yi'_iAaj. 

Neither  of  these  variables  is  of  the  type  to  which  the  Funda- 
mental Theorem  of  §  2  applies  ;  but  each  suggests  the  variable 

(15)  w(pc0  +  c)y0Ax-{-w(x1-{-c)ylAx-\ \-w(xn_1  +  c)yn.1£iX9 

whose  limit  is  the  definite  integral  (12),  and  each  approaches 
the  value  of  this  integral  as  its  limit,  as  we  will  now  show. 


166  CALCULUS 

Let        ak  =  w  (xk  +  c)  yk  Aaj,         Pk  =  w  (xk  +  c)  y[  Az. 

Then  lim(a14-«2+ \- 


o=/< 


w(x-\-c)ydx. 


Furthermore,     &  =  t£&±£}#jA*  =  |J  iim.^  =  1. 

a*      wfe  +  c^Aa      ?/A  n=»2/, 

Hence  /?i-f- A>  +  •••  +  /?«>  i.e.  the  variable  (13),  approaches  the 
value  of  the  above  integral  as  its  limit. 

In  like  manner  it  is  shown  that  (14)  approaches  this  same 
limit.  Hence  P  is  equal  to  this  limit  and  (12)  holds  in  all 
cases. 

7.  Length  of  a  Curve.  In  Chap.  VI  §  8  we  found  the  length 
of  a  curve  by  means  of  the  indefinite  integral, 


=jV'+(i: 


dx. 


We  can  also  evaluate  the 
required  length  by  con- 
sidering it  as  the  limit  of 
a  sum.  Divide  the  inter- 
val from  x  =  a  to  x  =  b 
into  n  equal  parts,  erect 
ordinates  at  the  points  of  division,  and  inscribe  a  broken  line 
in  the  arc  to  be  measured.     The  length  of  this  line  is 


Fig.  53 


X  VAa^  +  Aft* 


=  VAx2  +  A?/02  +  VAz2  +  A?/f  H f-  vAz2  +  A?/n_i2 


and  the  limit  of  its  length  is  the  length  s  to  be  determined. 
Now  this  latter  variable  is  not  of  a  type  whose  limit  is  a 
definite  integral,  but  it  suggests  a  new  sum  which  is  and 


DEFINITE  INTEGRALS  167 

whose  limit,  moreover,  can  be  identified  with  the  above  limit 
by  DuhamePs  Theorem  ;  namely,  since  lim  Ayk/  Ax  =f'(xk): 

VI  +f(xQfAx  +  VTTfW2Ax  +  •  •  •  +  Vl+/'K_1)2Aa;. 
For,  letting 

«* = vr+T7^?  ax,        pk=yji  + ^  a*, 

we  see  that  lim  &  =  1. 

n=oo   0^ 

Hence  5 

(16)  #=  fvi+f(xydx, 

a 

and  this  agrees  with  the  earlier  result  above  referred  to. 

8.  Area  of  a  Surface  of  Revolution.  To  find  the  lateral  area 
of  a  surface  of  revolution  we  proceed  in  a  manner  similar  to 
that  employed  in  finding  the  volume,  §  3.  Divide  the  interval 
from  x  =  a  to  x  =  b  into  n  equal  parts,  erect  ordinates,  and 
inscribe  a  broken  line  in  the  arc  of  the  generating  curve,  as 
in  the  preceding  paragraph.  This  broken  line,  when  it  rotates 
about  the  axis  of  revolution,  generates  the  lateral  surfaces  of  a 
series  of  frusta  of  cones.  Let  us  compute  the  lateral  area  of  the 
ft-th  frustum.  The  lateral  area  of  a  cone  is  half  the  product  of 
its  slant  height  by  the  perimeter  of  its  base,  -n-rl.  The  corre- 
sponding formula  for  the  frustum  is  the  product  of  the  slant 
height  by  the  circumference  of  the  circular  cross-section  made 
by  a  plane  passed  midway  between  the  bases, 

7r(r  +  M)l 

Hence  the  lateral  area  in  question  is 

(17)  v  (yk  +  Jfc+i)  VAz2  +  Ayk% 

and  the  area  of  the  surface  S  that  we  wish  to  compute  is  thus 
seen  to  be : 

(18)  8  =  lim  V  *  (yk  +  a* i)  Va^+a^7- 

n  =  oo  -<W 
*=0 


168  CALCULUS 


Now  the  general  summand  (17)  : 


ft  =  tt  (yk+1  +  yk)  ^1  +  ^L  As, 


Ax2 
suggests  the  simpler  expression  : 

%=27r2/AVl+/'(^)2Ax. 

The  sum  c^+c^-h  •••+«„  has  for  its  limit  I  2?r?/Vl+/'(»)ad.«. 

And  since  R  a 

lim^  =  l, 

«  =  oo   Ctk 

it  follows  from  Duhamel's  Theorem  that  the  sum  fSx  +  (32  +  •  •  • 
+  /?„,  has  the  same  limit.  But  the  limit  of  this  latter  sum  is, 
by  (18),  S.     Hence  we  obtain  the  result : 

6  ' 

(i9)  s=2,ryvi+f-i<to- 


For  example,  we  can  now  obtain  with  ease  the  theorem  of 
solid  geometry  that  the  area  of  a  zone  of  a  sphere  is  the  prod- 
uct of  its  altitude  by  the  circumference  of  a  great  circle, 
regardless  of  where  the  zone  is  situated.     We  have 

x2  +  y2  —  r2, 

1    ,   <¥_-,    ,  x2_r* 
dxr  y     y2 

a+h 

/r  a+h 

y-dx  —  2vrx        =  2  7T  rh,  q.  e.  d. 

y 

a 

The  area  of  the  complete  sphere  is  A-n-r2. 

EXERCISES 

1.  Find  the  area  of  a  segment  of  a  paraboloid  of  revolution, 
extending  from  the  vertex.  Ans.   §  it  ( Vm  (ra  +  2  xf  —  m2) . 

/  2.   Find  the  area  of  an  ellipsoid  of  revolution. 


>f>  Ans.   2tt(&2  + 


^sin^e 


DEFINITE   INTEGRALS  169 

3.  The  curve 

#J  H-  y  —  q> 

rotates  about  the  axis  of  x.     Find  the  total  superficial  area  of 
the  surface  generated.  .         12  -n-a2 

5 

4.  An  arch  of  a  cycloid  rotates  about  its  base.     Determine 
the  superficial  area  of  the  surface  generated.  ,        647rcr 

o 
*  5.   Show  that  in  polar  coordinates 

£  

(20)  ^2Wfritt^r»+  ££d§, 

a 

<  6.    Find  the  area  of  the  surface  generated  by  the  rotation  of 
the  cardioid 

r  =  2a(l  —  cos0) 

about  its  axis.  Ans.   +—  • 

5 

7.    Show  that  the  area  of  a  surface  of  revolution  is  given  by 
the  formula 


(21)  8 


=  2tt  I  yds, 


where  the  coordinates  x,  y  of  a  point  of  the  generating  curve 
are  expressed  as  functions  of  the  length  of  the  arc,  s. 

9.  Centre  of  Gravity.  A  Law  of  Statics.  Let  n  particles,  of 
masses  mlf  m2,  •••,  rnin,  be  situated  on  a  straight  line,  which  we 
will  take  as  the  axis  of  x,  and  let  their  coordinates  be  xl}  x2, 
-•-,xn.  Then  the  coordinate  x  of  their  centre  of  gravity  is  given 
by  the  formula  :  * 


(22) 


X 


—  mi ,Ti  ~f~  mzx2  4-  •  •  •  -f  m<nxT> 
m1  +  m<i+  ••-  +mn 


*  A  proof  of  this  formula  may  be  found  in  any  work  on  Mechanics,  for 
example,  Jeans,  Theoretical  Mechanics,  Chap.  VI. 


170  CALCULUS 

If  the  particles  are  situated  in  a  plane  at  the  points  (x1 ,  y^, 
•••,  0„,  y„)_or  in  space  at  the  points  (x1}  yu  z,),  ...,  (xn,  yn,  zn), 
and  if  (x,  y)  or  (x,  y,  z)  is  their  centre  of  gravity,  then  x  is 
given  by  the  same  formula  (22),  and  y  and  2  are  given  by 
similar  formulas,  in  which  the  cc's  are  all  replaced  by  2/'s  or  z's. 

Example.  A  granite  column  6  ft.  high  and  1%  ft.  in  diam- 
eter is  capped  by  a  ball  of  the  same  substance  2  ft.  in 
diameter  and  stands  on  a  cylindrical  granite  pedestal  9  in.  high 
and  2  ft.  in  diameter.  How  high  above  the  ground  is  the 
centre  of  gravity  of  the  whole  post  ?  Ans.  4.26  ft. 

10.  Centre  of  Gravity  of  Solids  and  Surfaces  of  Revolution. 

By  the  aid  of  the  Calculus  we  can  compute  the  centre  of 
gravity  of  bodies  not  made  up  of  a  finite  number  of  particles 
or  of  bodies  whose  centres  of  gravity  are  known.  We  will 
begin  with  homogeneous  solids  of  revolution.  Their  centre 
of  gravity  always  lies  somewhere  in  the  axis  of  symmetry. 
Divide  the  body  into  slabs  as  in  §  3,  Fig.  48,  and  denote  the 
abscissa  of  the  centre  of  gravity  of  the  k-th  slab  by  x£.  Then, 
if  p  denote  the  density  of  the  substance,  the  mass  of  the  k-th 
slab  is  pAVk,  and 

/3AF0  +  /3AF1+-"+/0AFn_1 


Here 


b 
V—tt  I  y2dx, 


and  it  remains  to  compute  the  value  of  the  numerator.  Since 
this  sum  has  the  same  value  for  all  values  of  n,  namely  x  V,  we 
may  allow  n  to  increase  without  limit,  and  we  shall  have 

(23)  Km  [x'0 A F0  +  x[  A  Vx  +  •  •  •  +  a£_i A  Vn-{\  =xV. 

n=oo 

Now  this  bracket  readily  suggests  a  sum  whose  limit  can  be 
computed  by  integration.     Since 


DEFINITE   INTEGRALS  171 

and  ^yl2Ax^  A  Vk^  iry'k'2Ax, 

where  y[  and  yH  are  respectively  the  smallest  and  the  largest 
radii  of  any  cross-section  of  the  &-th  slab,  we  see  that 

irxky'h2Ax <x'kAVk<  7rxk+1y'k,2Ax. 

Hence  if  we  put 

<*k  =  7r#*2/*2  A#,  ft  =  x'k  A  Vk, 

we  shall  have 

«&         #*       try k  AX  n=oo   aA 

It  follows,  then,  from  Duhamel's  Theorem  that  we  can  replace 
the  individual  terms  of  the  sum  in  (23),  namely  /3k  =  x'kAVk, 
"by  ak  =  7rxkyk2Ax.     We  thus  obtain: 

lim  [irx0y02Ax  +  irxlyfAx+ \-  nx^y^  Ax~]  =  x"V. 

n=oo 

The  left-hand  side  of  this  equation  is  a  definite  integral,  and 
so  we  are  led  to  the  result : 

7r  /  xy2dx 


(24)  x=. 


V 


For  example,  let  us  find  the  centre  of  gravity  of  a  cone  of 
revolution.     Here 

b  b 

/*  r2    /"  r2     x4 1  *      r2h2 

J xy2 dx  =  v, J **>=*■  l!o  =  T- 


and  K  =  l^r^  =  |A, 


i.e.  the  centre  of  gravity  is  three-fourths  of  the  distance  from 


172  CALCULUS 

EXERCISES 

1.   How  far  is  the  centre  of  gravity  of  a  hemisphere  from 
the  centre  of  the  sphere  ?  Ans.    §  r. 

*  2.   Find  the  centre  of  gravity  of  a  segment  of  a  paraboloid 
of  revolution. 

3.    Find  the  centre  of  gravity  of  a  segment  of  a  sphere. 

*  4.   Find  the  centre  of  gravity  of  a  frustum  of  a  cone. 

Ans.    Distance  from  smaller  base,     -  •    _^  "*"  _ — . 

4     R2  +  Rr  +  r2 

'  5.   The  curve 

y  =  sin  x,  0  ^  a?  ^  -  > 

rotates  about  the  axis  of  x.     Find  the  centre  of  gravity  of  the 
solid  generated.  ^    -=  *     1  =  ua 

4  7T 

«>  6.  Show  that  the  centre  of  gravity  of  a  surface  of  revolution 
bounded  by  two  planes  perpendicular  to  the  axis  is  giiien  by 
the  formula 


•w  I  xyds      2tt  I  ^2/A/l-f  -^dx 


H 


7.   Prove  that  the  centre  of  gravity  of  any  zone  of  a  sphere 
lies  midway  between  the  bases  of  the  zone. 

"  8.    Find  the  centre  of  gravity  of  the  lateral  surface  of  a  cone 
of  revolution.  Ans.   |7i. 

9.   Find  the  centre  of  gravity  of  the  lateral  surface  of  a 
segment  of  a  paraboloid. 

11.   Centre  of  Gravity  of  Plane  Areas.   To  find  the  abscissa 
of  the  centre  of  gravity  of  the  area  under  a  curve,  y  =f(x),  §  1, 
Fig.  47,  divide  the  area  into  n  >.*n-ips  of  equal  breadth  as  there 
■  .   li    I  eonsid"  i     '  .,  >■■■-->■.        ■     ,  : 

'    U  strip  be  denotet  ;      ..■..;..        ,  t^e^uper- 


i 


DEFINITE   INTEGRALS  173 

ficial  density,  supposed  constant  (i.e.  the  mass  of  one  square 
unit  of  the  slab),  by  p,  then  the  mass  of  the  A:-th  strip  will  be 
pAAk.  Let  the  abscissa  of  its  centre  of  gravity  be  x'k.  Then 
we  shall  have 

-  =  PAA  ■  a^  +  pAA  •  x[+  •••  +pA^n_1 .  xfn_x 
PAA0  +  p&A1+  ••'+pAAn_1 

_  x'0AA0  +  xl*Al+  •  •  •  +  <-i  AA-i 
A 

b 


Here 


jydx, 


and  it  remains  to  compute  the  value  of  the  numerator.  The 
reasoning  is  precisely  similar  to  that  of  the  preceding  para- 
graph.    We  allow  n  to  become  infinite  and  thus  obtain 

lim  [4 A Aq  +  x[  &AX  ■{ 4-  a?i_j  A An_{\  =  x A. 

Now  xk<x'k<xk+1, 

y'kAx^AAk^y'k'Ax, 

and  hence,  if        ak  =  xkykAx,        fik  =x'kAAk, 

lim&  =  l. 
w=ao  ak 

It  follows,  then,  from  Duhamel's  Theorem  that 

lim  [x0y0Ax+xly1Ax  +  •••  +  xH_1yn_lbx']=xAt 


b 

J- 


xydx 
(25)  ?~7Ti 

EXERCISES 
1.    Find  the  centre  of  gravity  of  a  semicircle. 


Ans.   x  =  —^  —  .425  r. 

07T 


174  CALCULUS 

2.  Find  the  centre  of  gravity  of  a  parabolic  segment. 

Ans.   x—^h. 

3.  Find  the  centre  of  gravity  of  half  an  ellipse,  bounded  by 

an  axis.  A   a    -      4a       ,0~ 

Ans.   x  =  —  =  .425a. 
Sir 

4.  Show  that  the  abscissa  of  the  centre  of  gravity  of  an 
arbitrary  plane  area  whose  boundary  is  cut  by  a  parallel  to  the 
axis  of  ordinates  at  most  in  two  points  is  given  by  the  formula : 

f  x(y"-y')dx 

*  =  °- A ' 

where  y'  =  <f>  (x)  is  the  equation  of  the  lower  boundary,  and  y" 
=f(x)  that  of  the  upper  one. 

5.  Show  that  the  centre  of  gravity  of  a  triangle  lies  at  the 
intersection  of  the  medians. 

6.  Show  that  the  centre  of  gravity  of  a  uniform  wire  of 
length  I  is  given  by  the  formula: 


/  x ds       j  #-%/ 


1+Xdx 


dx2 
x 


I  I 

7.   Find  the   centre  of   gravity  of   a  uniform  semi-circular 

wire.  A  2r      nnrr 

Ans.   x  —  —  =  .637r. 

7T 

12.  General  Formulation.  The  foregoing  examples  may  all 
be  brought  under  one  general  formulation,  which  applies 
furthermore  to  bodies  of  variable  density  and  wholly  arbitrary 
shape.  Let  the  body  be  divided  into  small  pieces,  and  denote 
the  mass  of  any  piece  by  £±Mk,  the  abscissa  of  its  centre  of 
gravity  by  x[.     Then  n-1 

M 


DEFINITE   INTEGRALS  175 


and  hence 

(26) 

a_— a 

M 

The  latter  limit  can  always  be  computed  by  means  of  integrals, 
but  it  may  be  necessary  to  employ  double  or  triple  integrals, 
cf.  the  later  chapters.  Formula  (26)  is  sometimes  written  in 
the  form : 

CxdM 

(27)  »-^r? 

13.  Centre  of  Fluid  Pressure.  Let  us  determine  at  what 
height  a  horizontal  brace  should  be  applied  to  hold  the  pressure 
of  the  liquid  computed  in  §  5,  without  there  being  any  tendency 
of  the  surface  to  rotate  about  the  brace.  Divide  the  surface 
into  strips,  as  in  §  5,  Fig.  51,  and  consider  the  pressures  on 
the  successive  strips.  As  in  the  problem  of  the  centre  of 
gravity  of  n  particles,  we  have  here  again  to  do  with  the  ab- 
scissa of  the  resultant  of  a  system  of  parallel  forces.  Let  the 
abscissa  of  the  point  at  which  the  pressure  &Pk  on  the  A;-th 
strip  acts  be  denoted  by  x[,  the  abscissa  of  the  brace  by  x. 
Then 

g  =  gjAP0  +  ariAP1+  -  +<-i±Pn-i 
APo  +  AA+.-.+AP^ 

The  value  of  the  denominator  is 


b 

P=w  I  (x-\-c)ydx. 


Allowing  n  in  the  numerator   to   increase  without  limit,  we 
obtain  by  reasoning  now  familiar  to  us  the  result : 


6 

?  I  (x-\-c)xydx 


176  CALCULUS 

For  example,  to  find  the  centre  of  pressure  for  the  rectangle 
considered  in  §  5.     Here 


b  h 


hs7c 


X  =  fA, 

and  the  brace  should  be  applied  to  the  end  of  the  tank  two- 
thirds  of  the  way  down. 

EXERCISE 

Find  the  depth  of  the  centre  of  pressure  in  the  case  of  the 
dam  described  in  §  5,  Ex.  1. 

14.  Moment  of  Inertia.  By  the  moment  of  inertia  of  a 
system  of  particles  about  a  straight  line  in  space,  called  the 
axis,  is  meant  the  quantity 

n 

(28)  ^  mkrk2  =  mvr?  +  m2r22  -\ \-  mnrn2, 

where  rk  denotes  the  perpendicular  distance  of  the  k-th.  particle, 
whose  mass  is  mk,  from  the  axis. 

If  a  body  consists  of  a  continuous  distribution  of  matter, 
like  a  wire  or  a  plate  or  a  solid  body,  its  moment  of  inertia  is 
defined  as  follows.  Let  the  body  be  divided  up  into  small 
pieces,  of  mass  mk,  and  let  the  mass  of  each  piece  be  concen- 
trated at  one  of  its  points,  whose  distance  from  the  axis  shall 
be  denoted  by  rk.  Form  the  sum  (28)  for  all  the  pieces. 
Then  the  limit  of  this  sum  as  the  pieces  grow  smaller  and 
smaller  is  the  moment  of  inertia  of  the  body  : 

(29)  /  =  limV  mfcrfc2. 

The  physical  meaning  of  the  moment  of  inertia  is  the 
measure  of  the  resistance  which  the  body  opposes,  through 


DEFINITE   INTEGRALS  177 

its  inertia,  to  being  rotated  about  the  axis.  The  moment  of 
inertia  also  has  an  important  application  in  the  theory  of  the 
strength  of  materials. 

By  means  of  the  Calculus  we  can  compute  the  moment  of 
inertia  of  any  body. 

Let  us  begin  with  a  circular  wire,  of  radius  r  and  mass  m, 
the  axis  being  perpendicular  to  the  plane  of  the  circle  and 
passing  through  its  centre.  Here  every  point  in  the  matter 
in  question  is  at  the  same  distance  r  from  the  axis,  and  so  the 
moment  of  inertia  is 

I=mr2. 

Next,  consider  a  uniform  circular  disc.  Divide  its  radius 
into  n  equal  parts:  r0  =  0,  ru  r2,  •••,  rn  =  a,  and  cut  the  disc  up 
into  rings  by  concentric  circles  of  radii  rlf  •••,  rn_x.  The  mo- 
ment of  inertia  of  the  whole  disc  is  equal  to  the  sum  of  the 
moments  of  inertia  of  these  rings.  Now  the  moment  of 
inertia  of  the  Axth  ring,  A7"x.,  evidently  is  greater  than  what 
it  would  be  if  its  mass  were  concentrated  along  its  inner 
boundary,  but  less  than  if  its  mass  were  concentrated  along 
its  outer  boundary.     Hence 

(30)  r,2A  Jf,  <  AJ,  <  r,+12 AMk. 

Furthermore,  AMk  =  pAAk}  where  p  denotes  the  density  and 
AAk  the  area : 

A.4,  =  7rr,+12  -  Trr,2  =  *(rk  +  Ar)2  -  Trr,2  =  2irr*Ar  +  tt  Ar2, 

(31)  AMk  =  2ttP  rk  Ar  +  irp  Ar2. 

We  are  now  in  a  position  to  apply  Duhamel's  Theorem.  We 
have 

I=AJ0-hAJi+...  -f-Ak., 

=  lim[A/e4-AJ1+-  4-Al^]. 

n=oo 

On  the  other  hand,  formulas  (30)  and  (31)  suggest  a  simpler 
infinitesimal  by  which  to  replace  A/A.,  namely 

ak=.27rprk3Ar. 


178  CALCULUS 

In  fact,  if  we  divide  (30)  through  by  ak : 


-k2AMk  Alk       ^WAJf, 


2  irpfj?  Ar      2  Trpr^.3  Ar     2  irprks  Ar 

,we  see  that  the  limit  of  either  extreme  is  1,  and  so  the  limit  of 
the  middle  expression  must  also  be  1.  Putting,  then,  f$k  =  AIk, 
we  get : 

n=oc    ak 

Hence 

a 

I  =  lim  V  2*prk**r  =  2vp  fr>dr  =  ^ . 

—  3  ^  2 

The  mass  of  the  disc  is  M  =  irpa2,  and  consequently  /may 
be  written  in  the  form : 
(32)  I  =  M°L. 

Definition.     If  the  moment  of  inertia  of  a  body  be  written  in 
the  form: 

I=Mk2, 

k  is  called  the  radius  of  gyration.  The  radius  of  gyration  is 
defined,  then,  as  s/I J M.  It  may  be  interpreted  as  follows : 
if  all  the  mass  were  spread  out  uniformly  along  a  circular  wire 
of  radius  k,  the  axis  passing  through  the  centre  of  the  ring  at 
right  angles  to  its  plane,  the  moment  of  inertia  would  still  be 
the  same :  /  =  M k2.  The  radius  of  gyration  of  the  above  cir- 
cular plate  is  a/  ^/2. 

EXERCISES 

Determine  the  following  moments  of  inertia. 

1.  A   uniform  rod,   of   length  2  a,   about  a  perpendicular 
bisector.  An^    Ma2 

2.  A  square  whose  sides  are  of  length  2  a,  about  a  parallel 
to  a  side  through  the  centre.  a         Ma2  ^ 

•  o 


DEFINITE  INTEGRALS  179 

3.   A  uniform  rod  of  length  I  about  a  perpendicular  through 
one  end.  *        Ml2 

3 

njr  2 

v4.   A  circular  disc  about  a  diameter.  Ans.   — —  • 

4 

/&.   An  isosceles  triangle  about  the  median  through  the  vertex. 

Ma2 
Ans.   — -f  where  a  is  half  the  length  of  the  base. 

6.    A  scalene  triangle  about  a  median. 

Mh2 

Ans.   ,  where  h  is  the  distance  of  either  vertex  from 

6 

the  median. 
J  7.   A  circular  wire  about  a  diameter. 


s  8.    A  cone  of  revolution  about  its  axis. 
9.   A  sphere  about  a  diameter. 


15.  A  General  Theorem.  When  the  moment  of  inertia  of  a 
body  about  an  axis  is  once  known,  its  moment  of  inertia  about 
any  parallel  axis  can  be  found  without  performing  a  new  inte- 
gration.    The  theorem  is  as  follows. 

Theorem.  If  the  moment  of  inertia  of  a  body  about  an 
arbitrary  axis  be  denoted  by  I0,  that  about  a  parallel  axis 
through  the  centre  of  gravity  by  i,  then 

(33)  I0  =  I+Mh\ 

where  h  denotes  the  distance  between  the  axes. 

We  will  prove  the  theorem  first  for  a  system  of  particles. 
Assume  a  set  of  cartesian  coordinates  (x,  y,  z),  the  axis  of  z 
being  taken  as  the  first  axis  of  the  theorem,  and  then  take  a 
second  set  of  cartesian  coordinates  (x',  y',  z')  parallel  to  the 
first,  the  origin  being  at  the  centre  of  gravity.     Then  we  have : 


Ans. 

Ma2 
2 

Ans. 

33fr2 
10 

Ans. 

2Ma2 

180  CALCULUS 

•4=2) wfS  =  2) m  (^2 + 2/2)> 

Furthermore, 

x=x'  +  x,         y  =  y'  +  y,         z  =  z'  +  z, 

where  (x,  y,  z)  is  the  centre  of  gravity  referred  to  the  (x,  y,  z) 
axes.     Hence 

2)  m(x*  +  y2)  =  2)  m (x'2  +  2/'2)  +2x2)  m*'  +  2?  2)  «? 

Now  2)  m^'  =  0>  2) Wl^' =  ^ 

For,  recall  formula  (22)  in  §  9.  Applying  that  formula  to  the 
present  system  of  particles,  referred  to  the  (V,  y',  2')-axes,  we 
see  that  the  abscissa  of  the  centre  of  gravity,  x\  is  : 

7  mx' 
M 

But  the  centre  of  gravity  is  at  the  new  origin  of  coordinates, 
and  so  x'  =  0,  hence  2)  mx>  =  0-     Similarly,  2jm.?/  =  0- 

It  remains  only  to  interpret  the  terms  that  are  left,  and 
thus  the  theorem  is  proven  for  a  system  of  particles. 

If  we  have  a  body  consisting  of  a  continuous  distribution  of 
matter,  we  divide  it  up  into  small  pieces,  concentrate  the  mass 
of  each  piece  at  its  centre  of  gravity,  form  the  above  sums, 

and  take  their  limits.  We  shall  have  as  before  2)mx'=0, 
V  my]  =  0,  and  hence 

2)  m(x2  +  2/2)  =  2)  m(aj'2  +  y'2)  +  Mli\ 

lim  2)  m(x2+if)=  lim  ^m(x'2  +  y'2)  +  Mh2,    q.  e.  d. 


DEFINITE   INTEGRALS 


181 


EXERCISES 


Determine  the  following  moments  of  inertia. 
1.   A  circular  disc  about  a  point  in  its  circumference.* 

Ans. 


SMa2 


2.   A  uniform  rod,  of  length  2  a,  about  a  point  in  its  perpen- 


dicular bisector. 


Ans.   Jlff^  +  h* 


3.   A   rectangle,   of  sides   2  a  and  26,  about  its  centre  of 
gravity.  A        M^2±^). 

o 


4.  The  following  figures 
about  the  axis  through  the 
centre  of  gravity  parallel 
to  the  lines  of  the  page  :         [^ 


in 


X 


D 


Fig.  54 

16.  The  Attraction  of  Gravitation.  Sir  Isaac  Newton  dis- 
covered the  law  of  universal  gravitation.  This  law  asserts 
that  any  two  particles  in  the  universe  attract  each  other  with 
a  force  proportional  to  their  masses  and  inversely  proportional 
to  the  square  of  the  distance  between  them: 


(34) 


/* 


mm' 


/-.*■ 


mm' 


where  K  is  a  physical  constant.f 

By  means  of  the  Calculus  we  can  compute  the  force  with 
which  bodies  consisting  of  a  continuous  distribution  of  matter 
attract  one  another.  Let  us  determine  the  force  which  a  uni- 
form rod  of  mass  M  exerts  on  a  particle  of  mass  m  situated 

*  By  the  moment  of  inertia  of  any  distribution  of  matter  in  a  plane 
about  a  point  in  that  plane  is  meant  the  moment  of  inertia  about  an  axis 
through  the  point  perpendicular  to  the  plane. 

t  Called  the  gravitational  constant.     Its  value  is 
6.5  x  10  8  cm3  sec~2gr-1. 


182  CALCULUS 

in  its  own  line.     Divide  the  rod  up  into  n  equal   parts   and 

denote  the  attraction  of  the  k-th  segment  by  AAk.     The  mass 

6  of   this    segment    is   pAx, 

;------ — -     I    '■"";.  ■'    '    '    '    H*..      where  p  denotes  the  den- 

x*       a  xo  x\  xk  «*+i  xn  r 

sity  of  the  rod.     Now  if 

FlG*  55  this  whole  mass  pAx  were 

concentrated  at  the  nearer  end,  its  attraction  would  be  greater 

than  AAk;  and  similarly,  if  it  were  concentrated  at  the  further 

end,  its  attraction  would  be  less.     Hence 

jr«e£f<.A4t<fSegs. 

^k+1  Xk 

It  follows,  then,  from  Duhamel's  Theorem,  if  we  set 


that 


-l 


A^^AA^hmJ^AA, 


Kmp 


_a     bj  ab 


Kmp  (b  —  a) 


This  result  may  be  written  in  the  form 

A=K^, 
ab 

and  thus  it  appears  that  the  rod  attracts  with  the  same  force 
as  a  particle  of  like  mass  situated  at  a  distance  from  m  equal 
to  the  geometric  mean  of  the  distances  a  and  b  of  the  ends  of 
the  rod. 

Secondly,  suppose  the  particle  were  situated  in  a  perpendic- 
ular bisector  of  the  rod.  Divide  the  rod  as  before  and  con- 
sider the  attraction  of  the  k-th.  segment.  We  must  now, 
however,  resolve  this  force  into  two  components,  one  perpen- 
dicular, the  other  parallel  to  the  rod.     The  latter  components 


DEFINITE  INTEGRALS  183 

annul  eacli  other  for  reasons  of  symmetry,  and  it  is  only  the 
sum  of  the  former, 

that  we  need  consider  further.  We  may  confine 
ourselves,  moreover,  to  half  the  rod  and  multiply 
the  final  result  by  2.     It  is  clear  that  * 

K^  cos  <f>k+1  <  AFk  <  K^cos  <f>k, 

and  hence  we  infer  by  the  usual  method  of  rea- 
soning that  „ 


*» 


Fig.  56 


Here  r2  =  h2  +  x2,  cos  <f>  =  —  , 

■Vh'  +  x2 
and  so  we  have 


F=2KmPh  f        dx 

J   V(ft2  +  ar>)3 


From  Peirce's  Tables,  Formula  138, 

dx  x 


f. 


V(^2+^)3      hWh2  +  x2 
and  consequently 

2Kmpa   _  ^      mM 
hVh2+a2  h^/h2  +  a2 


F=2Emp 


VW+x2 


EXERCISES 

Compute  the  following  attractions. 

1.    A  rod  whose  density  varies  as  the  distance  from  one  of 
its  ends,  on  a  particle  in  its  own  line. 

An8.  ?^M"i0gi±^+7A_-.r 


i2    &  h   i+h 


*  This  relation  holds  for  the  part  of  the  rod  we  are  considering,  namely, 
when  <fo>  0.    For  the  other  half  a  modification  is  necessary. 


184  CALCULUS 

t  2.   A  semicircular  wire,  on  a  particle  at  its  centre. 

Ans.    iS^f. 

^  3.   The  same  wire,  on  a  particle  in  the  circumference  situated 
symmetrically  as  regards  the  wire.  *        KmM^      .      3?r 

'*(**•  8  ' 

^  4.   A  rod  AB,  on  a  particle  situated  at  0  in  a  perpendicular 
O-B  at  one  end. 

Ans.   A  force  of  2KmM  sin  \AOB,   making    an   angle   of 
hi 

\AOB  with  OB. 
\l    5.   A  circular  disc,  on  a  particle  in  the  perpendicular  to  the 


disc  at  its  centre.  \        2KmM 


1- 


2  4- a2  J 


6.  A  rectangle,  on  a  particle  in  a  parallel  to  two  of  the  sides 
through  the  centre. 

For  further  simple  problems  in  attraction  cf.  Peirce,  New- 
tonian  Potential  Function. 

17.  Proof  of  Formula  (3).  We  can  give  a  proof  of  (3)  as 
follows.  Suppose  that  y  increases  with  x,  as  in  Fig.  57. 
Then  the  above  rectangles  are  all  inscribed  in  the  curve  and 
their  sum  is  less  than  the  area  A : 

(35)  /(a^Aaj+Z^Aa;  +  -  +/(av_I)Aa;<A 

Consider  now  a  second  set  of  rectangles  circumscribed  about 
the  strips  into  which  we  have  divided  A.  Their  sum  is 
greater  than  A : 

(36)  A  <f(x{)  Ax  +  •  •  •  +/(«V-i)  A*  +/(0  Ax. 

Thus  A  lies  between  the  two  variable  sums  (35)  and  (36). 
These  sums  differ  from  each  other  only  in  the  first  term  of 
(35),  /(a)  Ax,  and  the  last  term  of  (36),  / (b)  Ax,  i.e.  they 
differ  by  the  quantity  : 

(37)  [/(6)-/(a)]Ax, 


DEFINITE  INTEGRALS 


185 


a  quantity  that  approaches  0  as  its  limit  when  7i  =  cc.     Hence 
each  of  the  sums  (35)  and  (36)  approaches  A,  and  we  have  : 


(38) 


A=\im[f(x0)Ax+f(x1)bx+  ■■■  +f(x„_1)Ax] 
=  lim  [/(aO  Ax  +f(x2)  Ax  +  •  ••  +/(*»)  Ax    ]. 


A°) 


cr: 


~? 


en " 


just 


yw-A*) 


OCn-l    Xr?b 


Fig.  57 


Geometrically  the  difference  between  any  inscribed  rectangle 
and  its  corresponding  circumscribed  rectangle  is  the  area  of 
the  little  shaded  rectangle.  If  we  slide  all  these  latter  rectan- 
gles over  into  the  last 
strip,  they  will  form  a 
rectangle  whose  base  is 
Ax  and  whose  altitude 
isf(b)—f(a).  Its  area, 
then,  is  precisely  the. 
difference  (37). 

It  is  not  essential 
that  the  lengths  of  the  intervals  xl  —  x0  =  Ax0,  x2  —  xx  —  Axl9  ••• 
be  equal.  Let  the  greatest  of  these  lengths  be  denoted  by  h. 
Then  the  difference  between  the  sums 

f(x0)AxQ-j-f(xl)Axl+  •••  +f(xn_l)Axn_1, 

and  /(xL)  Ax{)  +f(x2  )Axl+ hf(xn )    Axn_i , 

as  is  seen  from  a  figure  similar  to  Fig.  57,  will  be  less  than 

and  so  each  sum  approaches  A  as  its  limit. 

If  y  decreases  as  x  increases,  the  reasoning  is  similar,  only 
the  sum  of  the  inscribed  rectangles  is  now  given  by  (36),  that 
of  the  circumscribed  rectangles  by  (35),  and  it  remains  to 
reverse  the  inequality  signs  in  (35)  and  (36)  and  change  the 
signs  in  (37). 

Finally,  if  the  curve  has  a  finite  number  of  maxima  and 
minima,  it  may  be  divided  into  segments  such  that,  in  each  of 
these,  y  steadily  increases  (or  remains  constant)  or  steadily 


186  CALCULUS 

decreases  (or  remains  constant).  That  part  of  the  total  sum 
which  corresponds  to  strips  lying  wholly  under  any  one  of 
these  segments  is  shown  as  in  the  preceding  discussion  to 
approach  the  area  under  that  segment  as  its  limit;  and  the 
sum  of  the  finite  number  of  terms  in  (35)  left  over  ap- 
proaches 0.  Hence  (3),  and  likewise  (5),  is  true,  even  when 
the  curve  has  a  finite  number  of  maxima  and  minima  in  the 
interval  (a,  b). 

Variable  Limits  of  Integration.  Let  f(x)  be  continuous  in 
the  interval  a  ^  x  <^  b,  and  let  x'  be  chosen  arbitrarily  in  this 
interval.     Then  the  definite  integral 


x' 


f(x)dx 


is  a  function  of  x',  <j>(x').  We  may  denote  the  variable  of  in- 
tegration, x,  by  t,  and  at  the  same  time  change  x'  to  x.  Thus 
we  have : 


(39)  *(*)  =  J/(0 


dt. 


The  integral  on  the  right-hand  side  represents  the  area 
under  the  curve,  bounded  by  a  variable  ordinate  whose  abscissa 
is  x.  Hence  its  derivative  has  the  value  (Chap.  VI,  §  1)  f(x), 
and  thus  we  see  that 

X 

(40)  *'(*)=/(*)        or         *.ff(f)dt=f(x). 

a 

Finally,  the  variable  of  integration,  t,  is  often  denoted  by  the 
same  letter  as  the  variable  upper  limit,  (40)  thus  being  written : 


X 

(41)  <f>(x)=Jf(x)dx. 


DEFINITE  INTEGRALS  187 


EXERCISES 


1.  A  cycloid   revolves   about    the    tangent  at  its  vertex. 
Show  that  the  volume  of  the  solid  generated  is  7r2a3. 

2.  Show  that  the  volume  of  the  solid  generated  by  a  curve 
that  revolves  about  the  axis  of  y  is  given  by  the  formula : 


V 


=  7T  I  xPdy. 


3.  A  cycloid  revolves  about  its  axis,  i.e.  the  line  through 
the  vertex  perpendicular  to  the  base.     Show  that  the  volume 

of  the  solid  generated  is  iro?l-^ )• 

4.  If  the  curve 

y2  (x  —  4  a)  =  ax  (x  —  3  a) 

revolve  about  the  axis  of  x,  show  that  the  volume  of  the  solid 
generated  by  the  loop  is  ?r  a3(7|  —  8  log  2).  Compute  this  vol- 
ume correct  to  three  significant  figures  when  a  —  1. 

Ans.   6.12. 

5.  The  curve  y2  —  x (x  —  1)  (x  —  2)  revolves  about  the  axis 
of  x.  Show  that  the  volume  of  the  solid  generated  by  the  oval 
is  \tt. 

6.  Find  the  volume  of  the  solid  generated  by  the  catenary 

y  =  i(e*  +  e-x) 
when  it  rotates  about  the  axis  of  y. 

Ans.   £[(r2-2r  +  2)er+(?*2  +  2r  +  2)e-r-4],   where  r  de- 

notes  the  radius  of  the  base. 

7.  Find  the  volume  of  a  torus  (anchor  ring).     Ans.   2irio2b. 

8.  Find  the  area  of  the  surface  of  a  torus. 

9.  Find  the  area  of  the  loop  of  the  curve 

Xs  —  3axy  +  ys  =  0.    .  Ans.   fa2. 


188  CALCULUS 

10.    Find  the  area  of  the  loop  of  the  curve 

r  cos  0  =  a  cos  20.  Ans.    (2--)a2. 


11.  Obtain  the  area  of  the  surface  of  a  segment  of  the  solid 
generated  by  the  rotation  of  a  catenary 

(a)  about  the  axis  of  x ;  (b)  about  the  axis  of  y. 

Ans.  (a)  7r(ys-\-ax)',  (b)  2  7r(a2-\-xs  —  ay),  where  s  denotes 
the  length  of  the  arc  measured  from  the  origin. 

12.  The  kinetic  energy  of  a  system  of  particles,  moving  in 
any  manner,  is  the  sum  of  the  kinetic  energies  of  the  individ- 
ual particles,  2^^mkvk2. 

Show  that  the  kinetic  energy  of  a  uniform  rod  of  length  2  a, 
which  is  rotating  about  its  perpendicular  bisector  with  angular 
velocity  o>,  is  i  Jfa2o>2. 

13.  A  pendulum  consisting  of  a  rectangular  lamina  oscillates 
about  an  axis  perpendicular  to  its  plane  and  passing  through 
the  middle  point  of  one  of  its  sides.  Compute  its  kinetic 
energy.  .  Ans.   M(±a?  +  \b2)oi2. 

14.  A  homogeneous  cylinder  rotates  about  its  axis.  Find 
its  kinetic  energy.  Ans.   \Md2o>2. 

15.  Show  that  the  kinetic  energy  of  a  rigid  body,  rotating 
with  angular  velocity  <o  about  any  axis,  is  JIco2,  where  I 
denotes  the  moment  of  inertia  about  the  axis. 

16.  The  density  of  water  under  a  pressure  of  p  atmospheres 
is  given  by  the  formula : 

p  =po  (1  +  .  00004  p), 


DEFINITE   INTEGRALS  189 

where  p  denotes  the  pressure  measured  in  atmospheres.  Show 
that  the  surface  of  an  ocean  six  miles  deep  lies  a  little  over 
600  ft.  deeper  than  it  would  if  water  were  incompressible. 

17.  The  perimeter  of  an  ellipse  whose  major  axis  2  a  is  twice 
as  long  as  the  minor  axis  can  be  shown  to  be  4.84  a.  {Infinite 
Series,  p.  30.)  Find  the  centre  of  gravity  of  a  uniform  wire 
in  the  form  of  half  such  an  ellipse,  the  ends  being  at  the 
extremities  of  the  minor  axis. 


CHAPTER  X 

MECHANICS 

1.  The  Laws  of  Motion.  Sir  Isaac  Newton  discovered  the 
laws  on  which  the  science  of  Mechanics  rests.  They  are  as 
follows : 

First  Law.  A  body  at  rest  remains  at  rest  and  a  body  in 
motion  moves  in  a  straight  line  with  unchanging  velocity,  unless 
some  external  force  acts  on  it. 

Second  Law.  The  rate  of  change  of  the  momentum  of  a 
body  is  proportional  to  the  resultant  external  force  that  acts  on 
the  body. 

Third  Law.     Action  and  reaction  are  equal  and  opposite. 

The  meaning  of  the  First  and  Third  Laws  is  obvious.  In 
the  Second  Law  the  momentum  of  the  body  is  to  be  understood 
as  the  product  of  its  mass  by  its  velocity,  mv.  And  since,  in 
the  vast  majority  of  cases  which  we  meet  in  practice,  the  mass 
is  constant,  we  have 

d(mv)==mdv 
dt  dt 

Now  the  rate  at  which  the  velocity  changes,  dv  /dt,  is  what  we 
commonly  call  acceleration,  —  we  will  denote  it  by  a; — and 
hence  the  Second  Law  may  be  expressed  as  follows : 

The  mass  times  the  acceleration  is  proportional  to  the  force  : 

(1)  ma  cc  /        or         ma  =  Xf 

190 


MECHANICS  191 

The  factor  A.  is  a  physical  constant.  Its  value  depends  on 
the  units  we  employ.  If  these  are  the  English  units:  foot, 
pound  (mass),  second,  and  pound  (weight),  X  has  the  value  32 : 
(2)  ma  =  32/. 

Furthermore,  since  v  =  —,  we  have : 
dt 

__  dv  _  d2s 

a~  dt~  dt2' 

In  applying  the  Second  Law  we  are  to  regard  a  force  which 

tends  to  increase  s  as  positive,  one  that  tends  to  decrease  s  as 

negative. 

If  forces  oblique  to  the  line  of  motion  *  act  on  the  body,  each 
one  must  be  broken  up  into  a  component  along  the  line  of 
motion  and  one  perpendicular  to  this  line.  The  latter  compo- 
nent has  no  influence  on  the  motion ;  the  former  component 
tends  to  produce  motion.  The  force  /  of  Newton's  Second  Law 
is  obtained,  when  several  forces  act  simultaneously,  as  the  al- 
gebraic sum  of  all  forces  and  components  of  forces  along  the  line 
of  motion,  taken  positive  when  they  tend  to  increase  s,  negative 
in  the  dther  case.  The  body  is  thought  of  as  moving  without 
rotation  and  may,  therefore,  be  conceived  as  a  particle. 

Finally,  we  will  deduce  a  new  expression  for  the  accelera- 
tion.    If  in  the  equation 

dv     dv  ds 

a  =  —  = , 

dt      ds  dt 

replace  ds/dt  by  its  value,  v.     Hence 

/Q\    .  dv 

(p)  a  =  v — 

v  J  ds 

Example  1.  A  freight  train  weighing  200  tons  is  drawn  by 
a  locomotive  that  exerts  a  pull  of  9  tons.  5  tons  of  this  force 
are  expended  in  overcoming  frictional  resistances.  How  much 
speed  will  the  train  have  acquired  at  the  end  of  a  minute  ? 

*  We  are  considering  here  only  the  case  of  rectilinear  motion  or  con- 
strained motion  in  a  given  curve.  For  a  more  general  statement  of  the 
law  cf.  §  9. 


192  CALCULUS 


p  f=mo  Here  we  have 

m  =  200  x  2000  =  400,000  lbs., 
/=  9  x  2000  -  5  x  2000  =  8000  lbs  * 


and  hence  (2)  becomes 

400,000— =  32x8000, 
dt 

dv     16 
or  —  =  — 

dt      25 
Integrating  with  respect  to  t,  we  get : 

v  =  i§t  +  a 

Since  v  =  0  when  t  =  0,  we  must  have  (7=0,  and  hence 

At  the  end  of  a  minute,  t  =  60,  and  so 

v  =if  x  60  =  38.4  ft.  per  sec, 
To  reduce  feet  per  second  to  miles  per  hour  it  is  convenient 
to  notice  that  30  miles  an  hour  is  equivalent  to  44  ft.  a  second, 
as  the  student  can  readily  verify ;  or,  roughly,  2  miles  an  hour 
corresponds  to  3  ft.  a  second.  Hence  the  speed  in  the  present 
case  is  two-thirds  of  38.4,  or  26  miles  an  hour. 

Example  2.     A  stone  is  sent  gliding  over  the  ice  with  an 
initial  velocity  of  30  ft.  a  sec.     If  the  coefficient  of  friction 
between  the  stone  and  the  ice  is  T^,  how  far  will  the  stone  go? 
f=-m  Here,  the  only  force  that  we  take 

—  '-  ^    *        ^ —     account  of  is  the  retarding  force  of 

friction,  and  this  amounts  to  one-tenth 
of  a  pound  of  force  for  every  pound 
of  mass  there  is  in  the  stone.  Hence,  if  there  are  m  pounds  of 
mass  in  the  stone  the  force  will  be  -fern  lbs.,t  an(i  since  it 
tends  to  decrease  s,  it  is  to  be  taken  as  negative : 

*  The  student  must  distinguish  carefully  between  the  two  meanings  of 
the  word  pound,  namely  (a)  a  mass,  and  (6)  a,  force;  —  two  totally  differ- 
ent physical  objects.  Thus  a  pound  of  lead  is  a  certain  quantity  of  matter. 
If  it  is  hung  up  by  a  string,  the  tension  in  the  string  is  a  pound  of  force. 

t  The  student  should  notice  that  m  is  neither  a  mass  nor  a  force,  but  a 
number,  like  all  the  other  letters  of  Algebra,  the  Calculus,  and  Physics. 


Fig.  59 


MECHANICS  193 


ma  =32 


(-fo) 


«=-¥• 

Now  what  we  want  is  a  relation  between  v  and  s,  for  the 
question  is  :  How  far  (s  =  ?),  when  the  stone  stops  (y  =  0).  So 
we  use  the  value  (3)  of  a : 

do  16 

or  vdv  =  —i^-ds. 

XT  D2  16       ,      , 

Hence  —  =  — —  s  +  C. 

£  o 

To  determine  O  we  have  the  data  that,  when  s  =  0,  v  =  30 : 

^-2  =  0  +  (7,  (7  =  450, 

v2  =  900--\*s. 
When  the  stone  stops,  v  =  0,  and  we  have 

0  =  900-^*,  s  =  141ft. 

EXERCISES* 

1.  An  ice  boat  that  weighs  1000  pounds  is  driven  by  a  wind 
which  exerts  a  force  of  35  pounds.  Find  how  fast  it  is  going 
at  the  end  of  30  seconds  if  it  starts  from  rest. 

Ans.  About  22  miles  an  hour. 

2.  A  small  boy  sees  a  slide  on  the  ice  ahead,  and  runs  for  it. 
He  reaches  it  with  a  speed  of  8  miles  an  hour  and  slides  15 
feet.  How  rough  are  his  shoes,  i.e.  what  is  the  coefficient  of 
friction  between  his  shoes  and  the  ice  ?  Ans.  fi  =  .15. 

3.  Show  that,  if  the  coefficient  of  friction  between  a  sprin- 
ter's shoes  and  the  track  is  T^,  his  best  possible  record  in  a 
hundred  yard  dash  cannot  be  less  than  15  seconds. 

*  The  student  is  expected  in  these  and  in  all  the  other  exercises  in  me- 
chanics to  draw  a  figure  for  each  exercise  and  to  mark  the  forces  distinctly 
in  it,  preferably  in  red  ink. 


194  CALCULUS 

4.  An  electric  car  weighing  12  tons  gets  up  a  speed  of  15 
miles  an  hour  in  10  seconds.  Find  the  average  force  that  acts 
on  it,  i.e.  the  constant  force  which  would  produce  the  same 
velocity  in  the  same  time. 

5.  In  the  preceding  problem,  assume  that  the  given  speed  is 
acquired  after  running  200  feet.  Find  the  time  required  and 
the  average  force. 

6.  A  train  weighing  500  tons  and  running  at  the  rate  of  30 
miles  an  hour  is  brought  to  rest  by  the  brakes  after  running 
600  feet.  While  it  is  being  stopped  it  passes  over  a  bridge. 
Find  the  force  with  which  the  bridge  pulls  on  its  anchorage. 

Arts.  25.2  tons. 

7.  An  electric  car  is  starting  on  an  icy  track.  The  wheels 
skid  and  it  takes  the  car  15  seconds  to  get  up  a  speed  of  two 
miles  an  hour.  Compute  the  coefficient  of  friction  between 
the  wheels  and  the  track. 

2.  Absolute  Units  of  Force.  The  units  in  terms  of  which 
we  measure  mass,  space,  time,  and  force  are  arbitrary.  If 
we  change  one  of  them  we  thereby  change  the  value  of  A.  in 
Newton's  Second  Law,  (1).  Consequently,  by  changing  the 
unit  of  force  properly,  the  units  of  mass,  space,  and  time 
being  held  fast,  we  can  make  X  =  1.     Hence  the 

Definition.  The  absolute  unit  of  force  is  that  unit  that 
makes  A.=  l  in  Newton's  Second  Law  of  Motion,  (1):* 

(4)  mo=/. 

*We  have  already  met  a  precisely  similar  question  twice  in  the 
Calculus.     In  differentiating  the  function  sin  x  we  obtain  the  formula 

Dx  sin  x  =  cos  x 
only  when  we  measure  angles  in  radians.     Otherwise  the  formula  reads 

Dx  sin  x  =  X  cos  x. 
In  particular,  if  the  unit  is  a  degree,  X  =  7r/180.     We  may,  therefore, 
define  a  radian  as  follows :     The  absolute  unit  of  angle  (the  radian)  is 
that  unit  that  makes  X  =  1  in  the  above  equation. 


MECHANICS  195 

In  order  to  determine  experimentally  the  absolute  unit  of 
force,  we  may  allow  a  body  to  fall  freely  and  observe  how- 
far  it  goes  in  a  known  time.  Let  the  number  g  be  the  number 
of  absolute  units  of  force  with  which  gravity  attracts  the 
unit  of  mass.  Then  the  force,  measured  in  absolute  units, 
with  which  gravity  attracts  a  body  of  m  units  of  mass  will 
be  mg.    Newton's  Second  Law  (4)  now  becomes : 

dv  ,  dv 

m-  =  mg,        hence  -  =  g; 

v=gt+C,  (7=0; 

ds 
V  =  dt=gt> 

s  =  igt*  +  K,  iT=0, 

and  we  have  the  law  for  freely  falling  bodies  deduced  directly 
from  Newton's  Second  Law  of  Motion,  the  hypothesis  being 
merely  that  the  force  of  gravity  is  constant.  Substituting  in 
the  last  equation  the  observed  values  s  =  S,  t  =  T,  we  get : 

9  fit 

if         jt2 

If  we  use  English  units  for  mass,  space,  and  time,  g  has, 
to  two  significant  figures,  the  value  32,  i.e.  the  absolute  unit 
of  force  in  this  system,  a  pounded,  is  equal  nearly  to  half  an 
ounce.  If  we  use  c.g.s.  units,  g  ranges  from  978  to  983  at 
different  parts  of  the  earth,  and  has  in  Cambridge  the  value 
980.  The  absolute  unit  of  force  in  this  system  is  called 
the  dyne. 

Since  g  is  equal  to  the  acceleration  with  which  a  body  falls 

Again,  in  differentiating  the  logarithm,  we  found 

Dx\ogax=(\ogae)-. 
x 

This  multiplier  reduces  to  unity  when  we  take  a  =  e.  Hence  the  defini- 
tion :  The '  absolute  (natural)  base  of  logarithms  is  that  base  which 
makes  the  multiplier  logae  in  the  above  equation  equal  to  unity. 


196 


CALCULUS 


freely  under  the  attraction  of  gravity,  g  is  called  the  accelera- 
tion of  gravity.  But  this  is  not  our  definition  of  g ;  it  is  a 
theorem  about  g  that  follows  from  Newton's  Second  Law  of 
Motion. 

The  student  can  now  readily  prove  the  following  theorem, 
which  is  often  taken  as  the  definition  of  the  absolute  unit 
of  force  in  elementary  physics:  The  absolute  unit  of  force 
is  that  force  which,  acting  on  the  unit  of  mass  for  the  unit 
of  time,  generates  the  unit  of  velocity. 

Example  1.  A  body  is  projected  down  an  inclined  plane  with 
an  initial  velocity  of  v0  feet  per  second.  Determine  the  motion 
completely. 

The  forces  which  act  are  :  the  component  of  gravity,  mg  sin  y 
absolute  units,  down  the  plane,  and  the  force  of  friction, 
fj,R  =  fxmg  cos  y  up  the  plane.     Hence 

ma  =  mg  sin  y  —  /xmg  cos  y 

dv 

or  —  s=  g  sin  y  —  fig  cos  y. 

at 

Integrating  this  equation,  we  get 

V  as  g  (sin  y  —  fx  COS  y)  t  +  C, 

v0=  0  +  C, 


(A) 


v=cti  =  9  ^sin  y-iMG0Sy)t  +  v<>- 


v0t, 


A  second  integration  gives 

(B)  8  =  \  g  (sin  y  —  fi  cos  y)  f 

the  constant  of  integration  here  being  0. 

To  find  v  in  terms  of  s  we  may  eliminate  t  between  (A)  and 
(B).     Or  we  can  begin  by  using  formula  (3)  for  the  acceleration : 

dv        .  .  v 

'U^":=9,(^Siny""/xC0S^' 

■|-'y2  =  gf(siny  —  ^,COSy)s     +  K, 

W=  0  +K, 

v2  =  2g  (sin  y  —  /x,  cos  y)  s  -f  v02. 


MECHANICS 


197 


Example  2.  An  Atwood's  machine  has  equal  weights,  M 
and  M ,  attached  to  the  cord,  and  a  rider  of  mass  m  is  added 
to  one  of  the  weights.     Determine  the  motion. 

We  apply  Newton's  Second  Law  to  each  of  the  weights 
M  and  M+m  individually.  The  forces  are  indicated  in  the 
diagram,  the  tension  in  the  string,  whose  weight  is  negligible, 
being  the  same  at  all  points.  Moreover,  since  the  space 
traversed  by  both  weights  is  the  same,  s, 
their  velocities   and   accelerations   are   also 

equal.     Thus 

M  a  =  T  -  Mg, 

(M  +  m)a=  (M  +  m)g-T, 


mg 
2M+m 


T  = 


(2M+27n)Mt 
2M+m       " 


Fig.  61 


From  the  last  formula  it  appears  that  the  tension  is  constant 
and  that  it  lies  between  the  values  Mg  and  (M+m)g.  The 
student  can  work  out  for  himself  the  formulas  that  give  v  and  s 
in  terms  of  t,  and  v  in  terms  of  s. 


EXERCISES* 

1.  A  weightless  cord  passes  over  a  smooth  pulley  and  car- 
ries weights  of  8  and  9  pounds  at  its  ends.  The  system  starts 
from  rest.  Find  how  far  the  9  pound  weight  will  descend 
before  it  has  acquired  a  velocity  of  one  foot  a  second.  What 
is  the  tension  in  the  cord  ?  Arts.   \\  ft. ;  8.4  lbs. 

2.  Obtain  the  usual  formulas  for  the  motion  of  a  body  pro- 
jected vertically : 

or         =-2gs+v02-, 
=  -gt  +  vQ\ 

=  -igt2+v0t. 

3.  On  the  surface  of  the  moon  a  pound  weighs  only  one  sixth 
as  much  as  on  the  surface  of  the  earth.     If  a  mouse  can  jump 


v2  =  2gs  +  v2 


gt  +  vQ 


or 


or 


See  foot  note  on  p.  193. 


198  CALCULUS 

up  1  foot  on  the  surface  of  the  earth,  how  high  could  it  jump 
on  the  surface  of  the  moon  ?  Compare  the  time  it  is  in  the  air 
in  the  two  cases. 

4.  A  bullet  fired  from  a  revolver  penetrates  a  block  of  wood 
to  a  distance  of  6  inches.  How  much  greater  would  its  velocity 
have  to  be  to  make  it  go  in  12  inches  ?  Assume  the  resistance 
to  be  the  same  at  all  points,  for  all  velocities. 

Ans.   About  40  percent. 

5.  Kegarding  the  big  locomotive  exhibited  at  the  World's 
Fair  in  1904  by  the  Baltimore  and  Ohio  Railroad  the  Scientific 
American  says :  "  Previous  to  sending  the  engine  to  St.  Louis, 
the  engine  was  tested  at  Schenectady,  where  she  took  a  63-car 
train  weighing  3,150  tons  up  a  one-per-cent.  grade." 

Find  how  long  it  would  take  the  engine  to  develop  a  speed 
of  15  m.  per  h.  in  the  same  train  on  the  level,  starting  from 
rest,  the  draw-bar  pull  being  assumed  to  be  the  same  as  on  the 
grade. 

6.  A  block  of  iron  weighing  100  pounds  rests  on  a  smooth 
table.  A  cord,  attached  to  the  iron,  runs  over  a  smooth  pulley 
at  the  edge  of  the  table  and  carries  a  weight  of  15  pounds, 
which  hangs  vertically.  The  system  is  released  with  the  iron 
10  feet  from  the  pulley.  How  long  will  it  be  before  the  iron 
reaches  the  pulley,  and  how  fast  will  it  be  moving  ? 

Ans.   2.19  sec. ;  9.1  ft.  a  sec. 

7.  Solve  the  same  problem  on  the  assumption  that  the  table 
is  rough,  /a  =2*0,  an(^  tnat  tne  Pu^ey  exerts  a  constant  re- 
tarding force  of  4  ounces. 

8.  If  Sir  Isaac  Newton  registered  170  pounds  on  a  spring 
balance  in  an  elevator  at  rest,  and  if,  when  the  elevator  was 
moving,  he  weighed  only  169  pounds,  what  inference  would  he 
draw  about  the  motion  of  the  elevator  ? 

9.  What  does  a  man  whose  weight  is  180  pounds  weigh  in 
an  elevator  that  is  descending  with  an  acceleration  of  2  feet 
per  second  per  second  ? 


MECHANICS  199 


(1 


10.  A  chest-weight  consists  of  a  movable  pulley 
and  a  fixed  pulley,  as  shown  in  the  diagram.  If  a  16 
pound  weight  is  attached  to  the  movable  pulley  and 
if  the  cord  carries  a  9  pound  weight  at  its  free  end, 
how  far  will  the  9  pound  weight  descend  before  it 
has  acquired  a  velocity  of  one  foot  a  second  ?  What 
is  the  tension  in  the  cord  ?  Ans.   J|  ft. ;  8.3  lbs.       Fl«-  62 

11.  In  a  system  of  pulleys  like  that  of  question  10  a  4  pound 
weight  is  attached  to  the  movable  pulley,  and  to  the  free  end 
of  the  cord  is  fastened  a  weight  of  1  pound  and  15  ounces, 
and  in  addition  a  rider  weighing  2  ounces  is  laid  on.  The  sys- 
tem starts  from  rest,  and  after  the  rider  has  descended  8  feet 
it  is  removed.     Determine  the  motion. 

12.  A  bucket  of  water,  at  the  bottom  of  which  there  rests  a 
stone,  forms  one  weight  of  an  Atwood's  machine.  The  bucket 
with  its  contents  weighs  16  pounds,  and  the  other  weight  is  18 
pounds.  If  the  stone  weighs  12  pounds  and  its  specific  gravity 
is  3,  find  how  hard  it  presses  on  the  bottom  of  the  bucket  when 
the  system  is  released. 

13.  In  the  bucket  described  in  the  preceding  question  there 
is  a  cork,  of  specific  gravity  \,  submerged  and  held  under  by  a 
thread  tied  to  the  bottom  of  the  bucket.  Will  the  tension  in 
the  thread  be  increased  or  diminished  after  the  system  is 
released  ? 

14.  What  is  the  mechanical  effect  on  one's  stomach  when 
one  is  in  an  elevator  which,  starting  from  rest,  is  allowed 
suddenly  to  descend? 

15.  A  block  of  ice  is  resting  on  a  sled,  the  coefficient  of  fric- 
tion between  the  ice  and  the  sled  being  ■£-$.  The  sled  is  drawn 
along,  starting  from  rest.  Find  the  shortest  possible  time  in 
which  the  ice  can  be  moved  10  ft. 

16.  A  man  weighing  180  pounds  is  at  the  top  of  a  building 
60  feet  above  the  ground.     He  has  a  rope  which  just  reaches 


200  CALCULUS 

to  the  ground  and  which  can  bear  a  strain  of  only  170  pounds. 
Can  he  slide  down  the  rope  to  the  ground  in  safety  ? 

Interpret  the  velocity  with  which  he  reaches  the  ground  by 
finding  the  height  from  which  he  would  have  to  drop  in  order 
to  acquire  the  same  velocity. 

17.  Find  the  shortest  time  in  which  a  bale  weighing  160 
pounds  can  be  raised  from  the  ground  to  a  window  25  feet  high 
(coming  to  rest  at  the  window)  by  means  of  the  rope  of  the 
preceding  question,  if  the  rope  passes  over  a  fixed  pulley  just 
above  the  window  and  is  drawn  in  over  the  drum  of  a  dummy 
engine. 

18.  If  the  speed  of  a  train  is  being  uniformly  retarded  by 
the  brakes,  prove  that  a  plumb  line  will  hang  at  rest  relatively 
to  the  train  at  a  certain  angle,  and  determine  this  angle. 

19.  In  the  train  described  in  the  preceding  exercises,  ques- 
tion 6,  there  is  a  bucket  of  water.  Find  the  angle  which  the 
surface  of  the  water  makes  with  the  plane  of  the  tracks  after 
the  water  has  ceased  to  surge. 

20.  At  what  angle  ought  a  man  to  stand  in  a  car  that  is 
starting  with  an  acceleration  of  3  feet  per  second  per  second  ? 

21.  The  drivers  of  a  locomotive  are  keyed  to  the  axle  and 
are  being  transported  on  a  platform  car.  The  axle  is  perpen- 
dicular to  the  track,  the  diameter  of  the  wheels  is  6  ft.,  and 
they  are  blocked  by  pieces  of  joist  3  in.  thick.  The  brakes 
being  put  on  hard,  so  that  the  train  loses  3. J-  miles  an  hour  of 
speed  every  second,  find  whether  the  drivers  will  jump  the 
cleats. 

22.  A  body  slides  down  a  smooth  inclined  plane.  Show 
that  the  velocity  with  which  it  reaches  the  foot  of  the  plane  is 
the  same  that  the  body  would  have  acquired  in  falling  freely 
through  the  same  difference  in  level. 

23.  Chords  are  drawn  from  the  highest  point  O  of  a  vertical 
circle.     Show  that  the  time  of  descent  of  a  bead  from  rest  at 


MECHANICS  201 

0,  down  a  smooth  wire  coinciding  with  any  one  of  these  chords, 
is  constant. 

24.  A  point  O  is  distant  10  feet  from  an  inclined  plane, 
whose  angle  of  inclination  is  a.  Find  the  shortest  time  in 
which  a  bead  can  reach  the  plane  if  it  starts  from  rest  at  0 
and  slides  down  a  smooth  straight  wire. 

25.  The  draw  bar  of  the  locomotive  in  Example  5  weighs 
50  pounds.  How  much  harder  does  the  engine  pull  on  the 
draw  bar  than  the  draw  bar  pulls  on  the  train  ? 

3.  Simple  Harmonic  Motion.  Problem.  One  end  of  an  elas- 
tic string  is  made  fast  at  a  point  A  and  to  the  other  end  is 
fastened  a  weight.  The  weight  is  carefully  brought  to  rest 
and  then  is  given  a  slight  vertical  displacement.  Determine 
the  motion. 

Let  AB  be  the  natural  length  of  the  string,  0  the 
point  of  equilibrium  of  the  weight,  and  let  P  be  the  , 
position  of  the  weight  at  any  instant  after  it  is  released; 
C,  the  point  from  which  it  is  released.  The  forces  that 
act  on  it  are :  the  force  of  gravity,  7ng,  downward  and 
the  tension  T  of  the  string  upward,  —  we  neglect  the 
damping  due  to  the  atmosphere.  Hence  we  have  from 
Newton's  Second  Law  of  Motion 

(5)  m-L=T-mg. 

From  Hooke's  law,  which  says  that  the  tension  in  a    o 
stretched  elastic  string  is  proportional  to  the  stretch- 
ing, it  follows  that  p  -- 

TIP 

(6)  t=^>  d 

where  X  is  Young's  modulus,*  provided  the  cross-section  of  the 

*  The  physical  constant  X  is  sometimes  interpreted  as  that  force  which 
would  be  required  to  double  the  length  of  the  string,  provided  this  could 
be  done  without  exceeding  the  elastic  limit. 


202  CALCULUS 

string  is  unity.     Since  at  0  the  tension  is  just  equal  to  the 
force  of  gravity,  we  have  furthermore 

(7)  mg  =  \22. 

Hence  from  (6)  and  (7) : 

and  thus  (5)  becomes 

(8)  mM  =  xT 

The  variables  s  and  x  are  connected  by  the  relation : 

s  +  x=OC=h, 

where  h  denotes  the  original  displacement.     Thus 

^£  .  ^  _  a  ds  _      dx 

~dt     ~dt~  °r         dt"      dt 

and  '^— *?• 

dt2  dt2 

Substituting  in  (8)  we  get 

d2x  =      X 
dt2  ml 

or,  setting  X/ml  =  n2 : 

This  differential  equation  is  characteristic  for  Simple  Harmonic 
Motion. 

To  integrate  (I)  multiply  through  by  2dx/dt  and  note  that* 


d  dx2 _r>dxd2x 
dt  dt2  ~    ~dt  dt2' 

have,  then : 

^dx  d2x _     2n2xdx 

dt  dt2  dt 

*  This  method  is  evidently  applicable  to  any  differential  equation  of  the 
form : 


MECHANICS  203 


Integrating  each  side  with  respect  to  t  we  get 

dx* 
dt2 


C-2n2x  —  dt  =  -2n2  Cxdx  =  -n2x>+C. 


To  determine  C  observe  that  initially  x  m  h,  while  the  velocity, 
equal  numerically  to  dx/dt,  is  0 : 

0  =  -n2h2  +  C. 
Hence 

(ii)  g.,*<»_rf). 

From  this  result  we  infer  (a)  that  the  maximum  velocity  is 
attained  when  x  =  0  and  is  ixh ;  (6)  that  the  height  to  which 
the  body  rises,  determined  by  putting  dx/dt  =  0  in  (II),  corre- 
sponds to  x  =  —  h.  The  latter  inference,  however,  is  legitimate 
only  on  the  assumption  that  the  point  C:  x  =  —  h,  is  not  higher 
than  B,  i.e.  that 

OC  =  OB. 

For  otherwise  the  body  will  rise  above  B,  and  since  the  string 
cannot  push,  a  new  law  of  force  becomes  operative,  the  force 
now  being  simply  that  of  gravity,  and  so  (I)  is  no  longer  true. 
We  return  to  equation  (II)  and  write  it  in  the  form 

dx 


^^-nVW-x2, 
dt 

the  minus  sign  holding  so  long  as  the  body  is  rising,  since  x 

decreases  as  t  increases.     To  integrate  this  equation  write  it 

as  follows: 

,  dx 

Hence 


nt 


J  VV-x2  h 


Initially  t  =  0  and  x  =  h,  therefore 

0=0  +  0 
and  we  have 


204  CALCULUS 

(9)  nt  =  cos"1-, 

hence 

(III)  x  =  h  cos  nt. 

We  have  deduced  this  result  merely  for  the  interval  that  the 
body  is  rising.  When  the  body  begins  to  descend,  dx/dt 
becomes  positive  and  we  have 

nt  =  +  I  _• 

J  Vh2-x2 

This  integral  can,  however,  still  be  expressed  by  the  formula 

(10)  w^cos-^+O, 

provided  that,  contrary  to  our  usual  agreement,  we  choose  that 
determination  of  the  multiple-valued  function  which  lies  be- 
tween 7r  and  2tt.  To  determine  Owe  have  from  (9)  that  when 
x  —  —h,  t  =  7r/n.     Substituting  these  values  in  (10)  we  get 

7r  =  cos-1(-l)+0,        (7  =  0, 

and  thus  (9)  and  (III)  hold  throughout  the  descent.  From 
this  point  on  the  motion  repeats  itself,  —  a  fact  that  is  mirrored 
analytically  in  equation  (III)  by  the  periodicity  of  the  function 
cos  nt.     Thus  formula  (III)  holds  without  restriction. 

Turning  now  to  a  detailed  discussion  of  these  results  we  see 
that  the  time  from  C  to  0  is  ^  =  ir/2n.  The  same  time  is  also 
required  from  0  to  C,  then  from  C  back  to  0,  and  lastly  from 
0  to  C.     Thus  the  total  time  from  C  back  to  C  is 

T=  —> 
n 

In  descending,  the  velocity  is  the  same  in  magnitude  as  when 
the  body  was  going  up,  only  reversed  in  sense ;  and  the  time 
required  to  descend  from  C  to  an  arbitrary  point  P  is  the 
same  as  that  required  to  rise  from  P  to  C. 

The  time  T  is  called  the  period  of  the  oscillation.  If  we 
consider  the  body  at  an  arbitrary  point  P  and  time  t,  then  at 


MECHANICS  205 

the  instant  T  seconds  later,  the  body  will  be  at  the  same  point 
and  moving  with  the  same  velocity,  both  in  magnitude  and 
sense,  —  this  fact  is  expressed  by  saying  that  the  phase  is  the 
same,  —  for 

x  =  hcosn(  H — —  )  =  h  cos  nt, 
\        n  J 

—  =  -hnsmn(t+  — ^  =  -  hn  sin  nt. 
dt  \        n) 

Finally,  we  observe  that  the  amplitude  2  h  of  the  oscillation 
has  no  effect  on  the  period. 

EXERCISES 

1.  One  end  of  an  elastic  string  is  fastened  at  a  point  A, 
and  to  the  other  end  is  attached  a  weight  that  would  just 
double  the  length  of  the  string.  The  weight  being  dropped 
from  A,  find  how  far  it  will  descend. 

Assume  the  string  to  be  3  feet  long  and  the  mass  of  the 
weight  to  be  2  pounds.  Ana.  11.2  ft. 

2.  If  the  weight  in  the  preceding  question  is  brought  to  a 
point  9  feet  below  A  and  released,  how  high  will  it  rise  ?  How 
long  will  it  take  for  it  to  return  to  the  starting  point  ? 

3.  A  slender  rod  is  clamped  at  one  end  so  as  to  be  horizontal 
when  not  loaded.  A  ball  of  lead  is  then  fastened  to  the  free 
end  and  brought  carefully  to  the  position  of  equilibrium, 
the  ball  dropping  by  less  than  3  %  of  the  length  of  the  rod. 
The  ball  being  given  a  slight  vertical  displacement,  show  that 
the  oscillation  will  be  approximately  simple  harmonic  motion 
and  determine  the  period. 

Neglect  the  deviation  of  the  path  of  the  ball  from  a  vertical 
straight  line,  and  assume  that  the  force  that  the  rod  exerts  is 
proportional  to  the  distance  which  the  free  end  has  been  dis- 
placed from  equilibrium. 

4.  A  steel  wire  of  one  square  millimeter  cross-section  is 
hung  up  in  Bunker  Hill  Monument,  and  a  weight  of  25  kilo- 


206  CALCULUS 

grammes  is  fastened  to  its  lower  end  and  carefully  brought  to 
rest.  The  weight  is  then  given  a  slight  vertical  displacement. 
Determine  the  period  of  the  oscillation. 

Given  that  the  force  required  to  double  the  length  of  the 
wire  is  21,000  kilogrammes,  and  that  the  length  of  the  wire  is 
210  feet.  Ans.   A  little  over  half  a  second. 

4.  Motion  under  the  Attraction  of  Gravitation.  Problem. 
To  find  the  velocity  which  a  stone  acquires  in  falling  to  the 
earth  from  interstellar  space. 

Assume  the  earth  to  be  at  rest  and  consider  only  the  force 
which  the  earth  exerts.  Let  r  be  the  distance  of  the  stone 
from  the  centre  O  of  the  earth,  and  s,  the  distance  it  has 
travelled  from  the  starting  point  A.  Then  the  force  acting  on 
it  is 

and  since  f—mg  when  r=R,  the  radius  of  the  earth: 

mg  =  —        and        /  =  — ^- — 
R2  T 

Hence,  from  Newton's  Second  Law  of  Motion, 


(11)  m 


d2s  _mgR2 


dt2         r2 

Furthermore,  s  +  r  =  l,  where  I  denotes  the  initial 
distance  OA,  and  consequently 

ds  ,dr_r\  d2s     d2r_Q 

Fig^64  dt     dt~    '  dt2      dt2 

Equation  (11)  tnus  becomes  : 

/  x  d2r         gR2 

To  integrate  this  equation,  we  employ  the  method  of  §  3  and 
multiply  by  2dr/dt: 

2dr<Pr=       2gR2dr 
" dt  dt2  r2    dt' 


MECHANICS  207 


Integrating  with  respect  to  t  we  get : 


Initially  dr/dt  =  0  and  r  =  I : 

Z 


0  =  ^  +  0,  C  = 


(5)  S^ffg-f 

Since  dr/dt  is  numerically  equal  to  the  velocity  ds/dt,  the 
velocity  V  at  the  surface  of  the  earth  is  given  by  the  equation : 

If  l  is  very  great,  the  last  term  in  the  parenthesis  is  small,  and 
so,  no  matter  how  great  I  is,  Fcan  never  quite  equal  ^s/2gR. 
Here  g  =  32,  72  =  4000  x  5280,  and  hence  the  velocity  in  ques- 
tion is  about  36,000  feet,  or  7  miles,  a  second. 

This  solution  neglects  the  retarding  effect  of  the  atmos- 
phere ;  but  as  the  atmosphere  is  very  rare  at  a  height  of  50  miles 
from  the  earth's  surface,  the  result  is  reliable  down  to  a  point 
comparatively  near  the  earth. 

In  order  to  find  the  time  it  would  take  the  stone  to  fall, 
write  (6)  in  the  form 

V/      rdr 


Hence  dt  = 


SB  ^/ir-r* 


and  t=-^i(     r^      . 

Turning  to  the  Tables,  No.  169,  we  find 

JVfr-r2  V  -Vlr- 


? 


208  CALCULUS 

=  _  VF^72  + \  sin"1  ^~1  +  K 

Thus       i- jj  {vs=*-J*r^}  +*. 

Initially  £  =  0  and  r  =  Z : 

8Ri        22  J 

Hence  finally : 

/•s  *     vTl    /i ■,,  i       _i2r-ZI 

(c)  ,»_jVF=?+j«o..»-f-J. 

EXERCISES 

1.  A  hole  is  bored  through  the  centre  of  the  earth  and  a 
stone  is  dropped  in.  Find  how  long  it  will  take  the  stone  to 
reach  the  centre  and  how  fast  it  will  be  going  when  it  gets 
there. 

Assume  that  the  air  has  been  exhausted  from  the  hole  and 
that  the  attraction  of  the  earth  is  proportional  to  the  distance 
from  the  centre. 

2.  Show  that  if  the  earth  were  without  an  atmosphere  and 
a  stone  were  projected  from  the  surface  of  the  earth  with  a 
velocity  of  ^2gB,  or  nearly  seven  miles  a  second,  it  would 
never  come  back. 

3.  The  moon's  mass  is  about  -^j  and  its  radius  about  T3T  that 
of  the  earth.  With  what  velocity  would  a  body  have  to  be 
projected  from  the  moon  in  order  not  to  return  ? 

4.  Taking  the  distance  of  the  moon  from  the  earth  as 
237,000  miles,  find  the  velocity  with  which  a  stone  would 
reach  the  moon  if  it  were  placed  at  the  point  of  no  force 
between  these  two  bodies  and  then  slightly  displaced  in  the 
direction  of  the  moon. 


MECHANICS  209 

5.  Find  how  long  it  would  take  Saturn  to  fall  to  the  sun. 
Given  that  the  acceleration  of  gravity  on  the  surface  of  the 
sun  is  905  feet  per  second  per  second,  that  the  diameter  of 
the  sun  is  860,000  miles,  and  that  the  distance  of  Saturn  from 
the  sun  is  880,000,000  miles. 

6.  How  long  would  it  take  the  earth  to  fall  to  the  sun? 
Given  that  the  distance  from  the  earth  to  the  sun  is  92,000,000 
miles. 

7.  How  long  would  it  take  the  moon  to  fall  to  the  earth  ? 

5.  Constrained  Motion.  If  a  particle  is  constrained  to  de- 
scribe a  given  path,  as  in  the  case,  for  example,  of  a  simple 
pendulum,  then  the  form  which  Newton's  Second  Law  of 
Motion  assumes  is  that  the  product  of  the  mass  by  the  acceler- 
ation along  the  path  is  equal  to  the  component,  along  the  path, 
of  the  resultant  of  all  the  forces  that  act. 

Consider  the  simple  pendulum.     Here 

d2s 
m  —  =  -mgsmd, 

and  since  s  =  10, 

(A)  «-{** 

This  differential  equation  is  characteristic  for  Sim- 
ple Pendulum  Motion.  We  can  obtain  a  first  integral 
by  the  method  of  §  3  : 

dt  dt2  I  dt' 


dl=-?l  fSm0d6  =  ^cos0  +  C, 
dtf  I  J  I 


0  =  ?2cosa+  G, 
i 


where  a  is  the  initial  angle ;  hence 

(B)  g  =  ^(0O8*-0O8«). 


210  CALCULUS 

The  velocity  in  the  path  at  the  lowest  point  is  I  times  the 
angular  velocity  for  0  =  0,  or  V2gl  (1  —  cos  a),  and  is  the  same 
that  would  have  been  acquired  if  the  bob  had  fallen  freely 
under  the  force  of  gravity  through  the  same  difference  in  level. 

If  we  attempt  to  obtain  the  time  by  integrating  (B),  we  are 
led  to  the  equation  : 

»_  JJ  f       *»  ■ 

*9j  Vcos0  —  cos  a 


This  integral  cannot  be  expressed  in  terms  of  the  functions  at 
present  at  our  disposal.  It  is  an  Elliptic  Integral.  When  6, 
however,  is  small,  sin#  differs  from  0  by  only  a  small  per- 
centage of  either  quantity,  Chap.  IV,  §  1,  and  hence  we  may 
expect  to  obtain  a  good  approximation  to  the  actual  motion  if 
we  replace  sin  6  in  (A)  by  6  : 

This  latter  equation  is  of  the  type  of  the  differential  equation 
of  Simple  Harmonic  Motion,  §  3,  (I),  n2  having  here  the  value 
g/l.  Hence  when  a  simple  pendulum  swings  through  a  small 
amplitude,  its  motion  is  approximately  harmonic  and  its  period 
is  approximately 


MwJ? 


9 

A  question  that  interested  the  mathematicians  of  the 
eighteenth  century  was  this  :  In  what  curve  should  a  pen- 
dulum swing  in  order  that  the  period  of  oscillation  may  be 
absolutely  independent  of  the  amplitude  ?  It  turns  out  that 
the  cycloid  has  this  property.  For  the  differential  equation  of 
motion  is 

m~=-mgsmT, 

where  s  is  measured  from  the  lowest  point,  and  since,  from 
Ex.  8,  p.  151, 


MECHANICS 


211 


we  have    — -  = 


Fig.  66 


5=  4a  sin  t, 

d2s  _       g_ 

~dX2~     4a  *' 
This  is  the  differential  equation 
of  Simple  Harmonic  Motion,  §  3, 
(I),  and  hence  the  period  of  the 
oscillation  : 

is  independent  of  the  amplitude. 

A  cycloid  pendulum  may  be  constructed  by  causing  the  cord 
of  the  pendulum  to  wind  on  the  evolute  of  the  path.  But  the 
resistances  due  to  the  stiffness  of  the  cord  as  it  winds  up  and 
unwinds  would  be  appreciable. 

We  will  close  this  paragraph  with  a  general  theorem.  Sup- 
pose a  bead  slides  on  a  smooth  wire  of  any  shape  whatever. 
Then  its  velocity  at  any  point  will  be  the  same  as  what  the 
bead  would  have  acquired  in  falling  freely  under  the  force  of 
gravity  the  same  difference  in  level. 

We  have  already  met  special  cases  of  this  theorem  in  the 
inclined  plane  and  the  simple  pendulum.  We  shall  restrict 
ourselves  to  plane  curves,  but  the  proof  can  be  extended  with- 
out difficulty  to  twisted  curves. 

Newton's  Second  Law  of  Motion  gives 

d2s  dx 

m-—  =  mgcosT=mg  —  t 

dfr  ds 

Hence     2*?**=2*  ^*  =  2<i^, 

dt  dt2        *  dsdt        y  dt 


df- 


=  2gx+C. 


(x0,y0) 


Fig.  67 


If  we  suppose  the  bead  to  start  from  rest  at  A,  then 
0  =  2gx0+C, 


(«) 


v*  =  —  =:2g(x-x0). 


212  CALCULUS 

But  the  velocity  that  a  body  falling  freely  a  distance  of  x  —  xQ 
attains  is  expressed  by  precisely  the  same  formula,  and  thus 
the  theorem  is  established. 

In  the  more  general  case  that  the  bead  passes  the  point  A 
with  a  velocity  v0  we  have  : 

v02  =  2gx0+C, 

(2T)  v2-v(?  =  2g(x-x0). 

Thus  it  is  seen  that  the  velocity  at  P  is  the  same  that  the  bead 
would  have  acquired  at  the  second  level  if  it  had  been  projected 
vertically  from  the  first  with  velocity  v0. 

The  theorem  also  asserts  that  the  sum  of  the  kinetic  and 
potential  energies  of  the  bead  is  constant,  or  that  the  change 
in  kinetic  energy  is  equal  to  the  work  done  on  the  bead. 

If  the  bead  starts  from  rest  at  A,  it  will  continue  to  slide 
till  it  reaches  the  end  of  the  wire  or  comes 

^L J_4'.     to  a  point  A'  at  the  same  level  as  A.     In 

the  latter  case  it  will  in  general  just  rise 
to  the  point  Af  and  then  retrace  its  path 
back  to  A.  But  if  the  tangent  to  the  curve 
at  A'  is  horizontal,  the  bead  may  approach 
A'  as  a  limiting  position  without  ever  reaching  it. 

EXERCISES 

1.  A  bead  slides  on  a  smooth  vertical  circle.  It  is  projected 
from  the  lowest  point  with  a  velocity  equal  to  that  which  it 
would  acquire  in  falling  from  rest  from  the  highest  point. 
Show  that  it  will  approach  the  highest  point  as  a  limit  which 
it  will  never  reach. 

2.  From  the  general  theorem  (51)  deduce  the  first  integral 
(B)  of  the  differential  equation  (A). 

6.  Motion  in  a  Resisting  Medium.  When  a  body  moves 
through  the  air  or  through  the  water,  these  media  oppose  re- 
sistance, the  magnitude  of  which  depends  on  the  velocity,  but 


MECHANICS  213 

does  not  follow  any  simple  mathematical  law.  For  low  veloci- 
ties up  to  5  or  10  miles  per  hour,  the  resistance  R  can  be 
expressed  approximately  by  the  formula : 

(12)  R  =  av, 

where  a  is  a  constant  depending  both  on  the  medium  and  on 
the  size  and  shape  of  the  body,  but  not  on  its  mass.  For  higher 
velocities  up  to  the  velocity  of  sound  (1082  ft.  a  sec.)  the 
formula 

(13)  R  =  cv2 

gives  a  sufficient  approximation  for  many  of  the  cases  that 
arise  in  practice.  We  shall  speak  of  other  formulas  at  the 
close  of  the  paragraph. 

Problem  1.  A  man  is  rowing  in  still  water  at  the  rate  of 
3  miles  an  hour,  when  he  ships  his  oars.  Determine  the  subse- 
quent motion  of  the  boat. 

Here  Newton's  Second  Law  gives  us: 

(14)  mTt  =  ~aV' 

a  v 

(15)  t  =  ™log% 

where  v0  is  the  initial  velocity,  nearly  4±-  ft.  a  sec. 
From  (15)  we  get : 

_at 

(16)  v=v0e  m. 

Hence  it  might  appear  that  the  boat  would  never  come  to  rest 
but  would  move  more  and  more  slowly,  since 

_  at 

lime  m  =  0. 

t  =  oo 

We  warn  the  student  strictly,  however,  against  such  a  conclu- 
sion.    For  the  approximation  we  are  using,  R  =  av,  holds  only 


214  CALCULUS 

for  a  limited  time  and  even  for  that  time  is  at  best  an  approxi- 
mation. It  will  probably  not  be  many  minutes  before  the 
boat  is  drifting  sidewise,  and  the  value  of  a  for  this  aspect  of 
the  boat  would  be  quite  different,  —  if  indeed  the  approxima- 
tion E  =  av  could  be  used  at  all. 

To  determine  the  distance  travelled,  we  have  from  (14) : 


and  consequently: 
(17) 


dv 
mv—  =  —  av, 


a 
m 


Hence,  even  if  the  above  law  of  resistance  held  up  to  the 
limit, *the  boat  would  not  travel  an  infinite  distance,  but  would 
approach  a  point  distant 

a 
feet  from  the  starting  point,  the  distance  traversed  thus  being 
proportional  to  the  initial  momentum. 

Finally,  to  get  a  relation  between  s  and  t,  integrate  (16)  : 

as  ~m 

dt=v°e  > 

(18)  s=^°(l-e~™). 

From  this  result  is  also  evident  that  the  boat  will  never  cover 
a  distance  of  8  ft.  while  the  above  approximation  lasts. 

EXERCISE 

If  the  man  and  the  boat  together  weigh  300  lbs.  and  if  a  steady 
force  of  3  lbs.  is  just  sufficient  to  maintain  a  speed  of  3  miles 
an  hour  in  still  water,  show  that  when  the  boat  has  gone  20  ft., 
the  speed  has  fallen  off  by  a  little  less  than  a  mile  an  hour. 

Problem  2.  A  drop  of  rain  falls  from  a  cloud  with  an  initial 
velocity  of  v0  ft.  a  sec.     Determine  the  motion. 

We  assume  that  the  drop  is  already  of  its  final  size,  —  not 


MECHANICS  215 

gathering  further  moisture  as  it  proceeds,  —  and  take  as  the 
law  of  resistance : 

R  =  cv2. 

Hence 


dv 

m—  = 

dt 

=  mg  —  cv2, 

ds 

mg  —  cv2 

i 
m 

ds  = 

mvdv 

mg  —  cv2 ' 

^-^log^-c^  +  C, 

0  =  -^log(mg-cv02)  +  C, 
and  thus  finally 

(19)  »  =  £logmg-CT5'- 

2  c       mg  —  cv2 

Solving  for  v  we  have 

^  _  mg  -  c  V 
m#  —  ctr 

(20)  <y2  =  ?M  _  ^  ~  C^Q2  e""£ 

c  c 

When  a  increases  indefinitely,  the  last  term  approaches  0  as 
its  limit,  and  hence  the  velocity  v  can  never  exceed  (or  quite 
equal)  v=  -y/mg/c  ft.  a  sec.  This  is  known  as  the  limiting 
velocity.  It  is  independent  of  the  height  and  also  of  the 
initial  velocity,  and  is  practically  attained  by  the  rain  as  it 
falls,  for  a  rain  drop  is  not  moving  sensibly  faster  when  it 
reaches  the  ground  than  it  was  at  the  top  of  a  high  building. 

EXERCISES 

1.  Show  that  if  a  charge  of  shot  be  fired  vertically  upward, 
it  will  return  with  a  velocity  about  3^  times  that  of  rain  drops 
of  the  same  size ;   and  that  if  it  be  fired  directly  downward 


216  CALCULUS 

from  a  balloon  two  miles  high,  the  velocity  will  not  be  appre- 
ciably greater. 

2.  Find  the  time  in  terms  of  the  velocity  and  the  velocity 
in  terms  of  the  time  in  Problem  2. 

3.  Determine  the  height  to  which  the  shot  will  rise  in  Ex.  1, 
and  show  that  the  time  to  the  highest  point  is 

£  =  -J— tan-M  v0\  —  ], 

where  v0  is  the  initial  velocity. 

7.  Graph  of  the  Resistance.  The  resistance  which  the  at- 
mosphere or  water  opposes  to  a  body  of  a  given  size  and  shape 
can  in  many  cases  be  determined  experimentally  with  a  reason- 
able degree  of  precision  and  thus  the  graph  of  the  resistance : 

JB-/60 

can  be  plotted.  The  mathematical  problem  then  presents  itself 
of  representing  the  curve  with  sufficient  accuracy  by  means  of 
a  simple  function  of  v.  In  the  problem  of 
vertical  motion  in  the  atmosphere, 

dv  ,   /./  N 

according  as  the  body  is  going  up  or  com- 
ing down,  s  being  measured  positively  downward.  Now  if  we 
approximate  tof(v)  by  means  of  a  quadratic  polynomial  or  a 
fractional  linear  function, 

a  +  ov  +  ckt        or         — '-£— , 
y  +  ov 

we  can  integrate  the  resulting  equation  readily.  And  it  is 
obvious  that  we  can  so  approximate,  —  at  least,  for  a  restricted 
range  of  values  for  v. 

Another  case  of  interest  is  that  in  which  the  resistance  of 
the  medium  is  the  only  force  that  acts : 

dv  j,/  N 


MECHANICS  217 

A  convenient  approximation  for  the  purposes  of  integration  is 

f(v)  =  avb. 

Here  a  and  b  are  merely  arbitrary  constants,  enabling  us  to 
impose  two  arbitrary  conditions  on  the  curve,  —  for  example, 
to  make  it  go  through  two  given  points,  —  and  are  to  be  deter- 
mined so  as  to  yield  a  good  approximation  to  the  physical  law. 
Sometimes  the  simple  values  6  =  1,  2,  3  can  be  used  with 
advantage.  But  we  must  not  confuse  these  approximate  for- 
mulas with  similarly  appearing  formulas  that  represent  exact 
physical  laws.  Thus,  in  geometry,  the  areas  of  similar  surfaces 
and  the  volumes  of  similar  solids  are  proportional  to  the  squares 
or  cubes  of  corresponding  linear  dimensions.  This  law  ex- 
presses a  fact  that  holds  to  the  finest  degree  of  accuracy  of 
which  physical  measurements  have  shown  themselves  to  be 
capable  and  with  no  restriction  whatever  on  the  size  of  the 
bodies.  But  the  law  R  =  av2  or  R  =  cy3  ceases  to  hold,  i.e. 
to  interpret  nature  within  the  limits  of  precision  of  physical 
measurements,  when  v  transcends  certain  restricted  limits,  and 
the  student  must  be  careful  to  bear  this  fact  in  mind. 


EXERCISES 

1.  Work  out  the  formulas  for  the  motion  of  the  body  in  each 
of  the  above  cases. 

2.  A  train  weighing  300  tons,  inclusive  of  the  locomotive, 
can  just  be  kept  in  motion  on  a  level  track  by  a  force  of  3 
pounds  to  the  ton.  The  locomotive  is  able  to  maintain  a  speed 
of  60  miles  an  hour,  the  horse  power  developed  being  reckoned 
as  1300.  Assuming  that  the  frictional  resistances  are  the 
same  at  high  speeds  as  at  low  ones  and  that  the  resistance  of 
the  air  is  proportional  to  the  square  of  the  velocity,  find  by 
how  much  the  speed  of  the  train  will  have  dropped  off  in 
running  half  a  mile  if  the  steam  is  cut  off  with  the  train  at 
full  speed. 


218  CALCULUS 

8.  Motion  under  an  Attractive  Force  with  Damping.    Let  us 

begin  with  a  concrete  example  and  consider  the  motion  of  the 
particle  of  §  3  when  the  resistance  of  the  atmosphere  is  taken 
into  account.  We  will  assume  that  this  force  is  proportional 
to  the  velocity,  =  —  kv.     Thus  (5)  is  replaced  by 

(21)  m^2=T-kv-mg, 
and  this  equation  becomes,  on  introducing  x : 

where   K  =  k/m,     n2  =  \/ml. 

Differential  equations  of  the  type  (21)  are  important  in 
physics.  They  occur  in  the  problem  of  the  damped  vibrations 
of  a  swinging  magnet,  but  especially  in  the  case  of  the  sus- 
pended coils  of  d'Arsonval  galvanometers.  One  method,  too, 
of  correcting  for  the  influence  of  the  atmosphere  on  the 
motion  of  a  pendulum  is  to  assume  (a)  that  the  moment  of 
inertia  is  slightly  increased,  i.e.  the  length  of  the  equivalent 
simple  pendulum  slightly  augmented,  and  (b)  that  the  resist- 
ance varies  as  the  velocity.  The  resulting  differential  equa- 
tion is  then  of  the  above  type. 

To  solve  a  differential  equation  is  to  find  a  function  which, 
when  substituted  in,  satisfies  the  equation.  By  the  order  of  a 
differential  equation  is  meant  the  order  of  the  highest  deriv- 
ative that  enters.  Thus  (2t)  is  of  the  second  order.  As  the 
general  solution  of  a  differential  equation  of  the  first  order 
we  expect  to  find  a  function  containing  one  arbitrary  con- 
stant ;  as  the  general  solution  of  a  differential  equation  of  the 
second  order,  a  function  containing  two  arbitrary  constants; 
and  so  on. 

In  order  to  solve  (2t)  we  make  use  of  an  artifice  and  inquire 
whether,  in  the  function 

(22)  x  =  emt, 

it  may  not  be  possible  so  to  determine  m  that  this  function 


MECHANICS  219 

shall  satisfy  (31).  (The  present  m  has,  of  course,  nothing  to 
do  with  the  earlier  m,  the  mass.)     Here 

G"K  mt  Of    33  2„mt 

—  as  me,  — -  =  me   , 
c&  dt2 

and  thus  the  left  hand  side  of  (51)  becomes,  on  substituting 

emt  f  0p  £, . 

e^fmH^  +  w2). 

Hence  we  see  that  if  m  is  chosen  as  either  one  of  the  roots 
of  the  quadratic  equation 

(23)  m2  +  Km  +  n*  =  0, 
i.e.  if                         m  =  —  %k±  Vj k2  —  w2, 

(51)  will  be  satisfied  by  (22).     Both  of  these  roots  are  negative, 
and  we  will  denote  them  by  —ml,  —  m2 ;  let  mx  <  ra2. 
More  generally,  the  function 

(24)  x  =  Ae-mit  +  Be~m*t 

also  satisfies  (2D,  as  is  shown  directly  by  substituting  in ;  and 
since  it  contains  two  arbitrary  constants,  it  is  the  general 
solution  of  (51)  for  the  case  that 

IV— *?>o. 

This  last  condition  would  not  be  fulfilled  in  the  case  of  §  3 
if  the  "  string "  were  a  steel  wire  and  the  weight  a  piece  of 
lead,  for  k  would  then  be  very  small.  It  could  be  realized, 
however,  if  the  "string"  is  a  spiral  spring  and  the  weight  is 
provided  with  a  collar,  to  act  like  an  inverted  parachute  and 
increase  the  damping.  To  determine  A  and  B  in  this  case  we 
have  that  initially  £  =  0,  x=h-,  hence 

(25)  h  =  A  +  B. 
Furthermore,  from  (24), 

—  =  -  mxAe-^x*  -  m^Be-™*, 
dt  2  , 

and  initially  dx/dt  =  0 ;  hence 


220  CALCULUS 

(26)  0  =  m1A-{-m2B. 

From  (25)  and  (26)  A  and  B  can  at  once  be  determined : 

a  __     m2h  p  _   —  mx  h 

ra2— -m/  m2  —  m1 

and  hence 

(27)  —  =■ - — - —  (e  "v  —  e  WV). 

dt  m2  —  w*i 

The  motion  is  now  completely  determined.  The  particle 
starts  from  rest  and  moves  upward  with  increasing  velocity 
for  a  time,  then  slows  up  and  approaches  the  point  «  =  0as  its 
limit,  when  t  =  oo  ,  —  practically,  of  course,  reaching  this  point 
after  a  comparatively  short  time.  All  this  we  read  off  from 
(24)  and  (27)  : 

lim  x  =  lim  (Ae-**  +  Be-™*' )  =  0  ; 

lim^  =  0; 

t=oo  dt 

0<t<oo, 
Fig.  70 

since  mx  <  ra2  and  consequently 

The  Case  \k2  —  n2  <  0.     If  on  the  other  hand 

(28)  iK2-n2<0, 

the  solution  (24)  becomes  illusory  through  the  presence  of 
imaginaries  in  the  exponents.  Now  in  the  algebra  of  im- 
aginaries 

e^~l  =  cos  <f>  +  V:rT  sin  <j>. 

Hence  (24)  becomes  : 


x=Ae  2  (cosVn2-iK2i+  V-lsinvV-iK2*) 


+  J3e  2    (cos  -Vn2  -^KH-V~1  sin  Vn2-i/<2  0, 
and  this  result  can  be  written  in  the  form 


MECHANICS  221 


(29)         x  =  e  2   (a  cos  V^2  -  i V « +  6  sin  Vn2  -  ^  *2  *), 

where  a  and  6  are  constants,  to  which  arbitrary  real  values 
can  be  assigned. 

The  foregoing  explanation,  by  means  of  imaginaries,  is  in 
no  wise  essential  to  the  validity  of  the  final  formula  (29). 
The  student  can  prove  directly  that  the  function  (29)  really 
is  a  solution,  no  matter  what  values  a  and  b  may  have,  by 
actually  substituting  it  in  (51). 

Another  form  in  which  the  solution  (29)  may  be  written  is 
the  following : 


(30)  x  =  Ce   2  sin  (Vn2-  \k*  t  +  y), 

where  C  and  y  are  now  the  constants  of  integration.  Instead 
of  the  sine  in  the  last  formula  the  cosine  may  equally  well  be 
written. 

Keturning  to  the  special  problem  before  us,  we  have,  for  the 
determination  of  C  and  y  in  (30),  initially :  x  =  h,  t  =  0: 

(31)  ft=<7siny. 
Furthermore, 

^  =  Ce"~s%  V»*  -J*1  cos  (Vw*  -  ±  #c»  1 4-  y) 


_!sin(Vn2-iK2*  +  y)], 
and  initially  dx/dt  =  0 : 

(32)  0  =  C\  Vri2-i*2cos  y  -  ^  sin  y  1 

From  (32)  it  follows  that 

COty: 


2Vn2-i*2 

If  we  take  the  solution  that  lies  in  the  first  quadrant 
0<y<^,  then,  from  (31),  C  will  be  positive,  and  we  shall 
have: 


222 


CALCULUS 


C=hcSGy  = 


7ih 


-Jn2-\K2 

The  accompanying  figure  represents  the  curve 
y=Ce-atsiii((3t+y), 

for  the  value  y  =  7r/4,  and  is  typical  for  the  whole  class  of 
curves  (30).     The  curves  cut  the  axis  of  abscissas  in  the  points 

. 7r  —  y  +  Tctt 


vV-!k2 


*=0,  1,2,..., 


and  hence  the  particle  passes  the  point  x  =  0  for  the  first  time 
(7r~  y)/Vw2  —  \k2  seconds  from  the  start,  and  continues  to  go 


s'"  Fig.  71 

through  this  point  periodically,  but  with  reversed  phase,  i.e.  in 
opposite  directions  at  intervals  of  7r/VV  — ^-*2  seconds;  with 
the  same  phase,  at  intervals  of 

rrr  27T 


^712-\k2 

seconds.     This   latter   quantity   is   called   the  period  of  the 
oscillation.     Since 

2?r        2?r      ttk*  .  f    terms  of  still     \ 

Vn2  —  i-K2       n       ^n3      \higher  order  in  *2/ 


MECHANICS  223 

as  will  be  shown  in  the  chapter  on  Taylor's  Theorem,  it  is 
seen  that,  when  k/u  is  small,  the  period  differs  bnt  slightly 
from  the  value 

n 

which  it  has  for  simple  harmonic  motion,  k  =  0.  The  effect 
of  the  damping  is  in  all  cases  to  lengthen  the  period. 

The  amplitude,  on  the  other  hand,  steadily  falls  off  toward 
0  as  its  limit  when  t  =  ao,  and  thus  the  particle  practically 
comes  to  rest  after  a  longer  or  shorter  time,  according  as  k/u 
is  small  or  comparatively  large.  But  so  long  as  the  oscillation 
is  perceptible,  the  period  is  the  same. 

The  Case  n2  —  \k2=  0.  Here  the  quadratic  (23)  has  equal 
roots,  and  thus  the  two  solutions 

e~m\l,  e_m2^ 

become  coincident.     7n1  =  m2  =  lK.     And  similarly, 

-** 


e    2  cosVw2  —  \k21         reduces  to         e    2 


while  e    2  sin  Vn2  —  \  k?  t 

vanishes  identically.  Thus  we  fail  to  get  a  solution  with  two 
arbitrary  constants  entering  in  such  a  way  that  we  can  impose 
two  independent  conditions  on  the  solution.  It  is  found  that 
in  this  case  the  general  solution  takes  the  form 

x  =  (D  +  Et)e~~2\ 

Determining  the  constants  D  and  E  as  in  the  cases  discussed 
above,  we  obtain : 


(33) 


(34)  ^.=  -^hte  ~2 


dt  4 


The  character  of  the  motion  is   the  same  as  in  the   case 


224  CALCULUS 

Jk2  — n2>0.  It  can,  however,  also  be  regarded  as  a  limiting 
case  under  n2  —  Jk2>0,  the  very  first  point  of  intersection  of 
the  curve  with  the  axis  of  abscissas  having  receded  to  infinity. 

9.  Motion  of  a  Projectile.  Problem.  To  find  the  path  of  a 
projectile  acted  on  only  by  the  force  of  gravity. 

The  degree  of  accuracy  of  the  approximation  to  the  true 
motion  obtained  in  the  following  solution  depends  on  the 
projectile  and  on  the  velocity  with  which  it  moves.  For  a 
cannon  ball  it  is  crude,  whereas  for  the  16  lb.  shot  used  in 
putting  the  shot  it  is  decidedly  good. 

Hitherto  we  have  known  the  path  of  the  body ;  here  we  do 
not.  We  may  state  Newton's  Second  Law  of  Motion  for  a 
plane  path  as  follows :  * 

d2x      ^r 

m^=Y, 
dt2 

where  X,  Fare  the  components  of  the  resultant  force  along 
the  axes. 

In  the  present  case  X=0,  Y=  —mg, 
and  we  have 

d2x     A 
dt2       ' 

d?y 
m — £  =  —  mg. 
dt2  y 


Fig.  72 


If  we  suppose  the  body  projected  from  0  with  velocity  v0  at 
an  angle  a  with  the  horizontal,  the  integration  of  these  equa- 
tions gives : 

•—  =  G  =  v0  cos  a,  x  =  v0t  cos  a ; 

etc 

*  The  form  of  Newton's  Second  Law  that  covers  all  cases,  both  in  the 
plane  and  in  space,  be  the  motion  constrained  or  free,  is  that  the  product 
of  the  mass  by  the  vector  acceleration  is  equal  to  the  vector  force. 


MECHANICS  225 


-1  =  v0  sin  a  —  gt,  y  =  vQt  sin  a  — %gt*. 

(XL 


Eliminating  t  we  get : 

(35)  y  =  x  tana  - 


2v02cos2a 
The  curve  has  a  maximum  at  the  point  A : 

vQ2  sin  a  cos  a  v2  sin2  a 

*>=^-l — >     *=v- 

Transforming  to  a  set  of  parallel  axes  through  A : 
a;  =  aj'  +  aj1,  y  =  y'  +  ylf 


we  find:  y<  =  -— -i* —  • 

2  V  cos  « 

This  curve  is  a  parabola  with  its  vertex  at  A.  The  height 
of  its  directrix  above  A  is  v02cos2a/2g,  and  hence  the  height 
of  the  directrix  of  (35)  above  0  is 

v02sm2a     v02cos2a_v02 
2g      +       2g      ~2g' 

This  result  is  independent  of  the  angle  of  elevation  a,  and  so 
it  appears  that  all  the  paths  traced  out  by  projectiles  leaving 
0  with  the  same  velocity  have  their  directrices  at  the  same 
level,  the  distance  of  this  level  above  0  being  the  height  to 
which  the  projectile  would  rise  if  shot  perpendicularly  upward. 

EXERCISES 
1.   Show  that  the  range  on  the  horizontal  is 


and  that  the  maximum  range  R  is  attained  when  a  =  45°: 

9 
The  height  of  the  directrix  above  0  is  half  this  latter  range. 


226  CALCULUS 

2.  A  projectile  is  launched  with  a  velocity  of  v0  ft.  a  sec. 
and  is  to  hit  a  mark  at  the  same  level  and  within  range.  Show 
that  there  are  two  possible  angles  of  elevation  and  that  one  is 
as  much  greater  than  45°  as  the  other  is  less. 

3.  Find  the  range  on  a  plane  inclined  at  an  angle  /?  to  the 
horizon  and  show  that  the  maximum  range  is 


Ra  = 


g  1  + sin/3 


4.  A  small  boy  can  throw  a  stone  100  ft.  on  the  level.  He 
is  on  top  of  a  house  40  ft.  high.  Show  that  he  can  throw 
the  stone  134  ft.  from  the  house.  Neglect  the  height  of  his 
hand  above  the  levels  in  question. 

5.  The  best  collegiate  record  for  putting  the  shot  is  46  ft. 
(F.  Beck,  Yale,  1903) ;  the  amateur  and  world's  record  is  49  ft. 
6  in.  (W.  W.  Coe,  Portland,  Ore.,  1905). 

If  a  man  puts  the  shot  46  ft.  and  the  shot  leaves  his  hand  at 
a  height  of  6  ft.  3  in.  above  the  ground,  find  the  velocity  with 
which  he  launches  it,  assuming  that  the  angle  of  elevation  a 
is  the  most  advantageous  one.  Ans.  vQ  =  35.87. 

6.  How  much  better  record  can  the  man  of  the  preceding 
question  make  than  a  shorter  man  of  equal  strength  and  skill, 
the  shot  leaving  the  latter's  hand  at  a  height  of  5  ft.  3  in.  ? 

7.  Show  that  it  is  possible  to  hit  a  mark  B :  (xb,yb),  provided 


Vb  4-  vV  +  yb- 

8.  A  revolver  can  give  a  bullet  a  muzzle  velocity  of  200  ft. 
a  sec.  Is  it  possible  to  hit  the  vane  on  a  church  spire  a  quarter 
of  a  mile  away,  the  height  of  the  spire  being  100  ft.  ? 


MECHANICS  227 


EXERCISES 

1.  A  cylindrical  spar  buoy  (specific  gravity  %)  is  anchored 
so  that  it  is  just  submerged  at  high  water.  If  the  cable  should 
break  at  high  tide,  show  that  the  spar  would  jump  entirely 
out  of  the  water. 

2.  A  number  of  iron  weights  are  attached  to  one  end  of  a 
long  round  wooden  spar,  so  that,  when  left  to  itself,  the  spar 
floats  vertically  in  water.  A  ten-kilogramme  weight  having  be- 
come accidentally  detached,  the  spar  is  seen  to  oscillate  with 
a  period  of  4  seconds.  The  radius  of  the  spar  is  10  centi- 
metres. Find  the  sum  of  the  weights  of  the  spar  and  attached 
iron.     Through  what  distance  does  the  spar  oscillate  ? 

Ans.    (a)  About  125  kilogrammes ;  (b)  0.64  metre. 

3.  A  chain  rests  partly  on  a  smooth  table,  a  piece  of  the 
chain  hanging  over  the  edge  of  the  table.  The  chain  being  re- 
leased, find  the  velocity  with  which  it  will  leave  the  table. 

4.  Solve  the  same  problem  for  a  rough  table,  the  chain 
passing  over  a  smooth  pulley  at  the  edge  of  the  table. 

5.  A  particle  of  mass  2  lbs.  lies  on  a  rough  horizontal  table, 
and  is  fastened  to  a  post  by  an  elastic  band  whose  unstretched 
length  is  10  inches.  The  coefficient  of  friction  is  \,  and  the 
band  is  doubled  in  length  by  hanging  it  vertically  with  the 
weight  at  its  lower  end.  If  the  particle  be  drawn  out  to  a 
distance  of  15  inches  from  the  post  and  then  projected  directly 
away  from  the  post  with  an  initial  velocity  of  5  ft.  a  sec,  find 
where  it  will  stop  for  good. 

6.  Show  that  if  two  spheres,  each  one  foot  in  diameter  and 
of  density  equal  to  the  earth's  mean  density  (specific  gravity 
5.6)  were  placed  with  their  surfaces  \  of  an  inch  apart  and 
were  acted  on  by  no  other  forces  than  their  mutual  attractions, 
they  would  come  together  in  about  five  minutes  and  a  half. 
Given  that  the  spheres  attract  as  if  all  their  mass  were  con- 
centrated at  their  centres. 


228  CALCULUS 

7.  A  particle  is  projected  horizontally  along  the  mner  sur- 
face of  a  smooth  vertical  tube.     Determine  its  motion. 

8.  A  man  and  a  parachute  weigh  150  pounds.  How  large 
must  the  parachute  be  that  the  man  may  trust  himself  to  it 
at  any  height,  if  25  ft.  a  sec.  is  a  safe  velocity  with  which  to 
reach  the  ground  ?  Given  that  the  resistance  of  the  air  is  as 
the  square  of  the  velocity  and  is  equal  to  2  pounds  per  square 
foot  of  opposing  surface  for  a  velocity  of  30  ft.  a  sec. 

Ans.   About  12  ft.  in  diameter. 

9.  A  toboggan  slide  of  constant  slope  is  a  quarter  of  a  mile 
long  and  has  a  fall  of  200  ft.  Assuming  that  the  coefficient 
of  friction  is  Tf -$,  that  the  resistance  of  the  air  is  proportional 
to  the  square  of  the  velocity  and  is  equal  to  2  pounds  per 
square  foot  of  opposing  surface  for  a  velocity  of  30  ft.  a 
sec,  that  a  loaded  toboggan  weighs  300  pounds  and  presents 
a  surface  of  3  sq.  ft.  to  the  resistance  of  the  air;  find  the 
velocity  acquired  during  the  descent  and  the  time  required  to 
reach  the  bottom. 

Find  the  limit  of  the  velocity  that  could  be  acquired  by  a 
toboggan  under  the  given  conditions  if  the  hill  were  of  infinite 
length. 

Ans.    (a)  68  ft.  a  sec. ;  (6)  30  sees. ;  (c)  74  ft.  a  sec. 

10.  The  ropes  of  an  elevator  break  and  the  elevator  falls 
without  obstruction  till  it  enters  an  air  chamber  at  the  bottom 
of  the  shaft.  The  elevator  weighs  2  tons  and  it  falls  from  a 
height  of  50  ft.  The  cross  section  of  the  well  is  6  x  6  ft.  and 
its  depth  is  12  ft.  If  no  air  escaped  from  the  well,  how  far 
would  the  elevator  sink  in?  What  would  be  the  maximum 
weight  of  a  man  of  170  pounds  ?  Given  that  the  pressure  and 
the  volume  of  air  when  compressed  without  gain  or  loss  of  heat 
follow  the  law : 

pv1A1  =  const., 

and  that  the  atmospheric  pressure  is  14  pounds  to  the  square 
inch. 


CHAPTER   XI 

THE  LAW  OF  THE   MEAN.    INDETERMINATE  FORMS 

1.  Rolle's  Theorem.  A  theorem  which  lies  at  the  founda- 
tion of  the  theoretical  development  of  the  Calculus  is  that  of 
Rolle,  from  which  follows  the  Law  of  the  Mean. 

Rolle's  Theorem.  If  <f>(x)  is  a  function  of  x,  continuous 
throughout  the  interval  a^x*£.b  and  vanishing  at  its  extremities : 

4>(a)  =  0,  <K&)=0, 

and  if  it  has  a  derivative,  —^r     =  <f>'(x),  at  every  interior  point 

ax 

of  the  interval,   then  <f>' (x)  must  vanish  for  at  least  one  point 

within  the  interval : 

<£'(X)=0,  a<X<b. 

For,  the  function  must  be  either  positive  or  negative  in  some 
parts  of  the  interval  if  we  exclude  the  special  case  that  <f>  (x) 
is  always  =  0,  for  which  case  the  y 
theorem  is  obviously  true.  Sup- 
pose, then,  that  <f>(x)  is  positive 
in  a  part  of  the  interval.  Then 
<f>(x)  will  have  a  maximum  at 
some  point  x  =  X  within  the  in- 
terval, and  at  this  point  the  derivative,  <f>'(x)  =  tan  t,  will  van- 
ish, cf.  Chap.  Ill,  §7: 

<£'(X)=0,  a<X<b. 

Similarly,  if  <f>(x)  is  negative,  it  will  have  a  minimum,  and 
thus  the  theorem  is  proven. 

229 


a  X 


*\ 


Fig.  73 


230 


CALCULUS 


2.  The  Law  of  the  Mean.  Let  the  function /(x)  be  contin- 
uous throughout  the  interval  a  ^  x  <^  b  and  let  it  have  a  deriv- 
ative, df(x)/dx=f,(x)}  at  every  interior  point  of  the  interval. 
Draw  the  graph  and  let  L M  be  the  secant  connecting  its  ex- 
tremities. Then  there  will  be  at  least  one  point  X  within  the 
interval  at  which  the  tangent  is  parallel  to  the  secant  LM. 

For,  consider  the  distance  from  a  point  P  of  the  curve  to 
the  secant,  measured  along  an  ordinate,  PQ.  This  distance 
(taken  algebraically)  will  have  a  maximum  or  a  minimum 
value,  and  at  such  a  point  the  tangent  is  evidently  parallel  to 
the  secant.     Now  the  slope  of  the  secant  is 

tan  ZJg£Jf=  A6} -/<«>, 

b  —  a 

A  )-Ka)  an(j  ^  si0pe  0f  the  curve  at 
x  =  X  is  /'(X).     Hence 

b—a 
(A)  /(&)-/<<*)  =  Q>  -  a)  f(X),  a<X<b. 

This  is  the  Law  of  the  Mean.  Another  form  in  which  it  is 
often  useful  to  write  the  theorem  is  obtained  by  setting 

b  —  a  =  h,  b  =  a-\-h. 

Then  X  can  be  written  as  a  -f-  Oh,  where  6  is  a  proper  frac- 
tion, or  at  least  a  positive  quantity  less  than  1,*  and  we  have : 

(A')  f(a  +  h)  -/(a)  =  hf  (a  +  6h),  0<6<1. 

In  (A),  a  and  b  can  be  interchanged  and  in  (A')  h  can  be 
negative. 

An  analytical  proof  of  the  Law  of  the  Mean  is  as  follows. 
Form  the  function 


+  {x) 


/(&) 


b  —  a 


m(x-a)~lf(x)-f(a)}. 


*  We  may  think  of  the  second  term,  dh,  as  representing  that  portion  of 
the  interval  b  —  a  —  h  which  must  be  added  to  the  segment  a  to  take  us 
toX 


LAW  OF   THE  MEAN.     INDETERMINATE   FORMS     231 

This  function  satisfies  all  the  conditions  of  Rolle's  Theorem 
and  hence  its  derivative, 

o  —  a 
must  vanish  for  a  value  x  =  X  between  a  and  b: 

/(*>)-/(<»)  _/(X)=0,  a<X<b. 

b  —  a 

Thus  the  theorem  is  proven. 

This  proof  merely  puts  into  analytic  form  the  geometric 
proof  first  given,  for  the  function  <f>(x)  here  employed  is  pre- 
cisely the  distance  PQ. 

3.  Application.  As  a  first  application  of  the  Law  of  the 
Mean  we  will  give  the  proof  of  Theorem  A  in  Chap.  VI,  §  2.  In 
that  theorem  3>'  (x)  =  0  by  hypothesis  for  all  values  of  x,  or 
at  least  for  all  in  a  certain  interval.  If,  then,  a  and  b  =  x1  are 
two  points  of  this  interval,  we  have  from  the  Law  of  the 
Mean  (A)  : 

<!>(x1)-&(a)  =  0, 

i.e.  ®(x1)  =  $(a)  for  all  points  xx  in  question.     Hence  &(x)  is 
a  constant. 

Exercise.  Show  that,  if  f(x)  satisfies  the  conditions  of  §  2 
and  if  furthermore  /'  (x)  >  0  at  all  points  within  the  interval, 
then 

4.  Indeterminate  Forms.  The  Limit  J.  If  both  the  numer- 
ator and  the  denominator  of  a  fraction 

(i)  /M 

vanish  for  a  particular  value  of  x,  x  =  a: 

/(a)=0,  2P(a)=0, 


232  CALCULUS 

the  fraction  takes  on  the  form  #  and  thus  ceases  to  have  any 
meaning.  The  fraction  will,  however,  in  general  approach  a 
limit  when  x  approaches  a,  and  we  proceed  to  determine 
this  limit. 

Sometimes  this  can  be  done  by  a  simple  transformation. 
Thus  if 

f(x)  _  x  —  a 
F(x)~ x2-a2' 

we  need  only  divide  numerator  and  denominator  by  x  —  a  and 
we  have : 

v      x  —  a       v         1  1 

lim  — =  lim sd  — . 

x=ax2  —  a2      x±ax  +  a     2a 

Again,  if  /M  =  *anx 

F(x)        x 
and  a  =  0,  we  have 

limtanf  =  Hm  J__    sinx  =  1_ 

x=o     x         x=o  cos  a;       x 

When,  however',  such  simple  devices  as  the  foregoing  are 
not  available,  we  can  apply  the  Law  of  the  Mean.  Let  b=x 
be  any  point  near  a.  Then,  remembering  that  f(a)  =  0  and 
F  (a)  =  0,  we  have : 

f(x)  =  (x-a)f'(X),  F{x)  =  (x-a)F'(X'), 

where  X  and  X  both  lie  between  a  and  x,  and  hence 

f(*y=r(X) 

F(x)      F'(X) 

When  x  approaches  a,  X  and  X  both  approach  a,  too,  and  so, 
if  f'(x)  and  F' (x)  are  continuous,  as  is  usually  the  case  in 
practice, 

lim/'(X)  =/'(a),  lim  F'  (X)  =  F'(o). 

If,  then,  F*  (a)  ^  0,  we  have : 

(2)  iim/©=/M 


LAW  OF  THE  MEAN.  INDETERMINATE  FORMS   233 

The  limit  of  such  a  fraction  as  the  one  above  considered  is 
referred  to  for  brevity  as  the  limit  $.* 

t?„„™^i„      rp„  i±„ri  •!•      logo; 


Here 


ue.     J. 

O  I1I1U    11 

x  = 

Ul  . 

=  1  1  —X 

/(*)= 

=  \ogx, 

/'<a»4 

/'(1)  = 

i; 

F(X): 

=  1—  X, 

F'(x)=*-1, 

«=1 1  —X 

1. 

F'(l)  = 

-i; 

EXERCISES 

Obtain  the  following  limits  without  differentiating. 

x-acc3  — a3      3a2  x=o  sinx 

2.  hm =  —  8.    Inn =  I. 

x=l    Xn  —  1         1\,  x=0         x 

3.  lim^=^  =  —  •  A     i^COS* 


lim 


ic3— a3     3a 
^  +  a_2     3 


m =  1. 

ZrCOtft 
2 

tan  ax 


i     tf-l         2  10.   lim""1  "*  =  <». 

..      ^  +  8       3 

5-  ii?,?T^=2o'  ii.  iimtafg!-Bma!»a 

x=0      1  —  COS  X 

'      e.    l^V^M-Va^J^^       i2     liml-cosx  =  l> 
*=o  x  2Va  x=o     sin2x         2 

13.    Obtain  the  limits  in  Exs.  2,  4,  7,9  by  differentiation. 

*  This  limit  is  also  called  the  "true  value"  of  the  "indeterminate 
form"  f(x)/F(x)  for  x  =  a.  Both  terms  are  based  on  a  false  conception. 
In  the  early  days  of  the  Calculus  mathematicians  thought  of  the  fraction 
as  really  having  a  value  when  x  =  a,  only  the  value  cannot  be  computed 
because  the  form  of  the  fraction  eludes  us.  This  is  wrong.  Division 
by  0  is  not  a  process  which  we  define  in  Algebra.  It  is  convenient,  how- 
ever, to  retain  the  term  indeterminate  form  as  applying  to  such  expres- 
sions as  the  above  and  others  considered  in  this  chapter,  which  for  a  cer- 
tain value  of  the  independent  variable  cease  to  have  a  meaning,  but  which 
approach  a  limit  when  the  independent  variable  converges  toward  the 
exceptional  value. 


234  CALCULUS 

Obtain  the  following  limits  by  differentiation. 

3=0         X  TV 


15.    lim- =  loga. 


•<     —4 


21.   Hm1-V2sip,ra;  =  -1, 


16.    lim^^!  =  log«.  -il-V2oos^x 

«*o      x  b  5  3 

22.   lim^-l+(x-l^=_3, 

o«  t     Va;  — Va      3   1 

-.«     r     e*  — e~x      n  23.  lira- —  =  -a?. 

18.  km — ; =  2.  ***</x--Va     2 

x=o    sin  a;  va      v« 

19.  lim  ^-Z^  =  _  ,.  24.    lim^  =  _-. 

x=-il  +  x  x=i2x  —  1  2 

oc     ,.     a4  +  3a,-3 -7a;2 -27  a; -18 

25*    IS  a4  -  3**  -  7s*  +  27a.  - 18".  ^  y<Wr  """"*• 

5.    A  More  General  Form  of  the  Law  of  the  Mean.    The 

method  of  evaluating  lim.  f(x)/F(x)  set  forth  in  §  4  is  inap- 
plicable when  /'  (a)  and  F'  (a)  both  vanish,  for  then  /'  (a)/F'(a) 
ceases  to  have  a  meaning.  Moreover,  since  we  do  not  know 
how  X  and  X'  vary,  —  it  is  not  at  present  clear  that  they  can 
be  taken  equal  to  each  other, — we  cannot  see  what  limit 
f'(X)/F'(X')  approaches.  We  can  deal  with  this  and  other 
cases  that  arise  by  the  aid  of  the  following 

Generalized  Law  of  the  Mean.  Iff(x)  and  F(x)  are  con- 
tinuous throughout  the  interval  a  <  x  <  b  and  each  has  a  deriva- 
tive at  all  interior  points  of  the  interval,  and  if,  moreover,  the 
derivative  F'  (x)  does  not  vanish  within  the  interval;  then,  for 
some  value  x  ==  X  within  this  interval, 

K   >  F(b)-F(a)     F'(X)' 


LAW  OF  THE  MEAN.  INDETERMINATE  FORMS  235 

The  proof  is  as  follows.     Form  the  function : 

This  function  satisfies  all  the  conditions  of  Rolle's  Theorem, 
and  hence  its  derivative, 

must  vanish  for  a  value  of  x  within  the  interval.     Hence 

F^F%F,(-X)-f{X)  =  0'  a<X<b- 

By  hypothesis,  F'  (x)  is  never  0  in  the  interval.  Consequently 
we  are  justified  in  dividing  through  by  it,  and  thus  we  obtain 
Formula  (B),  q.e.d. 

6.  The  Limit  - ,  Concluded.  We  can  now  state  a  more  gen- 
eral rule  for  determining  the  limit  considered  in  §  4.  Suppose 
f(a)  =0  and  F(a)  =  0.  Let  a;  be  a  point  near  a  and  set  b  =  x 
in  (B).     Then 

F(x)     F'(X)' 

where  now  we  have  the  same  X  in  both  numerator  and  denomi- 
nator, and  X  lies  between  x  and  a.  When  x  approaches  a,  X 
will  also  approach  a.  Hence,  if  f'(x)/F'(x)  approaches  a 
limit,  f  (X)  /F'  (X)  will  approach  the  same  limit,  and  so  will 
its  equal,  f(x)/F{x).     Thus  we  have : 

(i)  iim./M=iim^. 

V    '  x±aF(x)         x±aF'(x) 

If,  then,  it  turns  out  on  differentiating  that  f  (a)  =  0  and 
F'  (a)  =  0,  we  can  differentiate  again,  and  so  on. 

Example.   To  find      lim  ^—^ — ~sm  ■« 
*=o     1  —  cos  x 


236  CALCULUS 

Here  /(*)  =  <?*  -  cos  a 

F'(x)  sin  a; 

and  the  new  ratio  is  still  indeterminate  when  x  =  0.  Differ- 
entiating again  we  have 

lim!l±smx  =  1_ 

x=o      cos  a; 
Hence  the  value  of  the  original  limit  is  1. 

7.   The  Limit  §-.     The  rule  for  finding  the  limit  of  the  ratio 

(I)  when  both  the  numerator  and  the  denominator  become 
infinite  for  x  =  a : 

/(a)  =  oo,  F(a)  =  oo, 

is  the  same  as  when  both  the  numerator  and  the  denominator 
vanish,  namely  :  Differentiate  the  numerator  for  a  new  numera- 
tor, the  denominator  for  a  new  denominator,  and  take  the  limit 
of  the  new  ratio : 

(II)  lim-^)=limm. 

7  x=aF{x)         x=aF'(x) 

To  prove  this  theorem  let  us  first  take  the  case  that  a  =  oo  , 
i.e.  that  the  independent  variable  x  increases  without  limit. 
In  the  Generalized  Law  of  the  Mean  (B),  replace  a  by  x'  and  b 

(L*  /(*)-/&)     f(X)  x'<X<x 

(tJ)  F(x)-F(x')-FVT)'  *<X<*, 

and  write  the  left-hand  side  in  the  form : 

a)  fV).  l~f(x')/f(x) 

v  J  Fix)    1-F(x')/  F(x) 

It  is  easily  seen  that  the  second  factor,  which  we  will  denote 

by  A: 

l-f(x')/f(x)  _ 

1-F(x')/F{*)'~* 

can  be  made  to  approach  1  as  its  limit.  For,  as  x  and  x'  in- 
crease without  limit,  both  f(x)  and  f(x'),  and  also  F(x)  and 
F(x'),  become  infinite.     Now  x  and  x'  are  independent  of  each 


LAW  OF   THE   MEAN.    INDETERMINATE   FORMS      237 

other.  We  may,  therefore,  choose  x'  so  that,  while  still  becom- 
ing infinite  as  x  becomes  infinite,  it  increases  so  much  more 
slowly  than  x  that 

f(x)  F(x) 

On  the  other  hand,  X  always  lying  between  x'  and  x  and  there- 
fore becoming  infinite  with  them,  it  is  clear  that,  if  /'  (x)  /F'  (x) 
approaches  a  limit  when  x  =  ao,  then  f'(X)/F'(X)  will  ap- 
proach the  same  limit.  Hence,  writing  (3)  by  the  aid  of  (4) 
in  the  form : 

f(x)  _lf'(X) 

F(x)      \F'(Xy 

we  see  that  the  right-hand  side  approaches  as  its  limit  the 
same  limit  that  f'(x)  /F'(x)  approaches.  The  left-hand  side 
*must,  therefore,  also  approach  this  limit,  and  the  theorem  is 
proveohr  when  a  —  oo  . 

If  x  approaches  a  limit  a,  we  need  only  to  introduce  a  new 
variable : 

1  ,  1 

x  —  a  y 

Setting    f(x)=f(a  +  ^J  =  <t>(y),         F(x)  =F(a  +  ^  =  <P(y), 

we  have  from  the  foregoing  result : 

But  *'(y)  =/(*)  z£  f(y)  =  F'(x)rl, 

y  y 

4>'(y)  =  f(*) 

*'(y)      F'{x)       ■ 

If',  then,  f'(x)/F'(x)  approaches  a  limit  when  x  approaches 
a,  <f>' (y)  / ^' (y)  will   approach   the   same   limit  when  y=cc. 


238  <  CALCULUS 

Hence  <f>(y)/&  (y)  will  approach  this  limit,  too.     But  <£  (y) /<£  (y) 
=f(x)/F(x).     This  completes  the  proof. * 

Example.     To  find  lim—. 

*=«  ex 

We  have :  lim  -  ==  lim  -  =  0. 


8.  The  Limit  0  •  oo .  If  we  have  the  product  of  two  func- 
tions : 

f(x)  .<£(>), 

one  of  which  approaches  0  as  x  approaches  a,  while  the  other 
becomes  infinite,  we  can  determine  the  limit  of  this  product 
by  throwing  the  latter  into  one  of  the  forms : 

m    or    m, 

i.e.  the  form  0/0  or  oo/oo,  and  then  applying  the  foregoing 
methods. 

Example.     To  find  lim  x logic. 
Here  it  is  better  to  choose  the  form 

1/x 
for  then  the  logarithm  will  disappear  on  differentiation : 

limlES*    lim-l^L  m  iim  (_  x)  =  o. 

x=o  x~L       x=o  —  1/ar       x±o 

*  The  theorem  contained  in  (2)  goes  back  to  1' Hospital,  1696.  The 
theorem  of  this  paragraph  is  due  to  Cauchy,  1823  and  1829,  who  proved 
it,  however,  only  on  the  assumption  that  f(x)  /  F(x)  approaches  a  limit. 
Stolz  extended  it  in  1879  as  in  the  text,  showing  that,  if  /'  (x)  /  F'  (x) 
approaches  a  limit,  then /(a:)  /  F(x)  will  also  approach  a  limit,  and  this 
will  be  the  same  limit. 


LAW  OF  THE   MEAN.    INDETERMINATE   FORMS     239 

EXERCISES 

Determine  the  following  limits. 

2 

1.  lim--  Ans.  0.        6.   lim  x  log  sin  x.        Ans.  0. 

*=«  ex  *=° 

2.  lim?".  Ans.  0.        7.    lim-^iL  ^tw.  3. 
3=00  ex                                                i=o  cot  3x 

3.  lim  a;  cot  trx.  Ans.—        8.   lim£calogx,  a>0.  ^4ws.  0. 

x  =  0  7T  x=0 

4.  iim?2gi?.  ^4rts.  0.        9.    lim-^-.  .4ws.  oo. 

i  =  oo         X  2  =  oo  l0gi(7 

B.  liml°££,»>0.     Aw.  0.     10.  iimlogsin2a;.       J,*.! 

x=oo    aw  x=o  log  sin  a 

9.   The  Limits  0°,  l00,  go0,  and  oo  —  oo.     The  expression 
(5)  /(s)*<*> 

ceases  to  have  a  meaning  when/(#)  and  <£(#)  take  on  certain 
pairs  of  values.     If  we  write  (cf.  Formula  (5)  on  p.  77) 

f(x)  =  elog/(x)  fix)^^  =  e*(x)  los/(-x^> 

we  see  that  the  expression  (5)  becomes  indeterminate  when  the 
exponent  of  e  takes  on  the  form  0-oo.  We  are  thus  led  to 
consider  new  limits  of  the  types  : 

(a)  /(a)  =  0,  *(a)=0;  0°. 

(b)  /(a)  =  l,  *(o)  =  oo;  1". 

(c)  /(a)  =  »,  *(a)=0;  oo0. 

The  limiting  value  of  the  exponent  of  e.  can  be  obtained  by 
the  method  of  §  8,  and  hence  the  limit  of  (5)  determined. 

Example.   To  find  lim  (cos  x)*3. 

I  log  COS  X 

(cos  x)x3  =  e    *    , 

limlogCosx  =  lim   -sinx. 
x=o       a?  x=o3arcosa? 


240  CALCULUS 

This  last  limit  can  be  obtained  immediately  by  a  simple  trans- 
formation : 

—  sin  x  _  1  sin  x 

o  x2  cos  x         3x  cos  x       x 

Hence   we  see   that   the   exponent   of   c   becomes  negatively 
infinite  if  x  approaches  0  from  the  positive  side,  and  so 

1 

3  =  0     V  ' 

If,  however,  x  approaches  0  from  the  negative  side,  the  ex- 
ponent of  e  becomes  positively  infinite,  and 

j_ 

lim  (cos#)*3=oo. 

z=0 

A  convenient  notation  for  distinguishing  between   these  two 
cases  is  the  following  : 

_L  J_ 

lim  (cos  x)x3=  0,  lim  (cos  x)x3—  oo. 

x=0+  »  =  0- 

TJie  Limit  co  —  oo .    If  we  have  the  difference  of  two  functions, 
each  of  which  is  becoming  infinite,  as 

log  (x  + 1)  —  log  x 

when  x  =  oc,  it  is  sometimes  possible  to  evaluate  the  limit  by  a 
simple  transformation.     For  example  : 

log  (x  + 1)  -  log  x  =  log  (l  + 1\  lim  log  (*  +  -Y=  0. 

More  often,  however,  the  simplest  method  is  that  of  infinite 
series,  cf.  Chap.  XIII. 

EXERCISES 

Determine  the  following  limits. 

-i-  1 

1.    lim  of .  Arts.    1.      3.    lim#1-a:#  Ans.    -■ 

x  =  0  x=l  6 


2.    lim(l  +  sina;)cotz.  Ans.    e.     4.    lim(Vl  +  X2— a;).   ^4ns.   0. 


LAW   OF   THE   MEAN.     INDETERMINATE   FORMS      241 


5.  Km  (cot  x)x.  Ans.   1.     ft     ,.     /    , 

fa        V  •*f^ 

6.  limf -  +  1  ]  •  Ans.  ea. 

*=«\x        J 

JL  1 

7.  lim  (cos  a;)*2.        Ans.    — .    1Q     H 


^4?is.    —  1. 


8.  iiai[2-.-Y"«- 


log  a?      log  a?y 

^4ns.    —  1. 


EXERCISES 

Determine  the  following  limits. 


1.    lim 


2-3x  +  4:x5 


*  =  *  Ix  +  tf  +  lx5 

o      r  3  4-"  a? 

2.  lim  — -• 

x  =*  4  —  9  a?  4-  a;2 

3.  IimV9W, 


11.    lim 


12.    lim 


13.    lim  sin  a;  (log  a;)2. 


x=«  (x  —  a)'' 


14.    lim" 


5.    lim 


I ,     .  15.    lim  csc2/3x  log  cos  aa;. 

a  —  x     a  4- a?  J  *-° 

6.    limvY?^2cot^J^.    16.     lim**'"1' 


7.    lim 


cos-1  a; 


»*»  Vl-a;2 


7ra; 


a?sina? 


8.    lim 


17.  lim  (1  —  a?)  tan 

z  =  l  ii 

18.  lima~xloga?. 


cos  a; 


x 


19.    lim 


a?2  — a? 


9.    lim  n  sin  - 

n  =  »  71 

10.    lim—. 

x  =  «   x3 


*=»1  —  a;  4- log  a; 

»/3— — 


20.    lim 


l_a  +  2Vl4-a-24-a;4 


242 


2i.    lim}*— * -J*- 

22.    lim  csc  #  sin  (tan  as). 


CALCULUS 

23.    lim  xa(logxy,  a>0,  £>0. 


24.    lira 


Qog^y 


m  >  0,  n  >  0. 


25.     lim  G(x)e~x,  where  G(x)  is  a  polynomial. 


26.    Show  that 


p      X* 

lim— =  0, 


n  being  any  constant  whatever. 


CHAPTER   XII 

CONVERGENCE  OF  INFINITE   SERIES* 

1.   The   Geometric   Series.     We  have  met  in   Algebra  the 
Geometric  Progression : 

a-t-ar  +  ar2-\ , 

the  snm  of  the  first  n  terms  of  which  is  given  by  the  formula: 

a  —arn 
i  1  —  r 

Suppose,  for  example,  that  a  =  1,  r  =  1.     Then 

Sl  =  l  =1 

etc. 

If  we  plot  on  a  line  the  points  which  represent  sr,  s2,  s3,  •••, 
it  is  easy  to  see  how  to  obtain  sn  from  its  predecessor,  sn_u 

Si  Sn    So 

.  —\ 1 — I ^-A4 

Fig.  75 

namely  :  sn  lies  half  way  between  sn_j  and  the  point  2.  Hence 
it  appears  that,  when  n  grows  larger  and  larger  without  limit, 
sn  approaches  2  as  its  limit. 

*  This  chapter  is  in  substance  a  reproduction  of   Chapter  I  of  the 
author's  Introduction  to  Infinite  Series,  published  by  Harvard  University. 
^  243 


244  CALCULUS 

In  general,  if  r  is  numerically  less  than  1, 

|'r|<l,         i.e.         -l<r<l, 
rn  will  approach  0  as  its  limit  when  n  —  oo,  and  we  shall  have : 

lim  sn  = • . 


We  have  here  an  example  of  an  infinite  series,  whose  value 
is  a/(l  —  r)  : 

(1)  ^L_  =  0-far4-<w*+-,         |r|<l, 
1  —  r 

and  we  turn  now  to  the  general  definition  of  such  series. 

2.   Definition  of  an  Infinite  Series.     Let  w0,  ulf  u2,  •••  be  any 

set  of  values,  positive  or  negative  at  pleasure.     Form  the  sum  : 

(2)  sn  =  u0  +  ux  +    •   •   •   4-  un_x. 

When  n  increases  without  limit,  sn  may  approach  a  limit,  U: 

lim  sn  =  U. 

In  this,  case  the  series  which  stands  on  the  right-hand  side  of 

(2)  is  said  to  converge  and  to  have  the  value  U*     It  is  cus- 
tomary to  express  both  of  these  facts  by  the  equation : 

(3)  U=u0  +  Ul+   .  •   -. 

But  if  sn  approaches  no  limit,  the  series  is  said  to  diverge. 

Such  a  series  is  called  an  infinite  series.  An  infinite  series, 
then,  is  a  variable  consisting  of  the  sum  of  n  terms. f  It  is 
said  to  be  convergent  if  the  value  of  this  sum,  sn,  approaches 
a  limit  when  n  =  oo  ;  otherwise  to  be  divergent.  And  in  the 
case  of  convergence  its  value  is  defined  as  lim  sn.  No  value 
is  assigned  to  a  divergent  series. 

*  £7 is  often  called  the  "sum"  of  the  series.  But  the  student  must 
not  forget  that  IT  is  not  a  sum,  but  is  the  limit  of  a  sum.  Similarly,  the 
expression,  "the  sum  of  an  infinite  number  of  terms"  means  the  limit 
of  the  sum  of  n  of  these  terms,  as  n  increases  without  limit. 

t  Each,  term  of  the  series,  however,  as  w0  or  u\  or  w*,  is  independent 
of  the  number  of  terms  n  involved  in  the  above  sum. 


CONVERGENCE   OF   INFINITE   SERIES  245 

Examples  of  divergent  series  are : 

1  +  2  +  3  +  4+    •-, 

1-1  +  1-1+  .... 
A  notation  commonly  employed  for  the  series  (3)  is 

Y?{B         or,  more  explicitly :       Vv 

n=0 

Thus  the  geometric  series  (1)  would  be  written : 

oo 

X: 


=  o 


3.  Tests  for  Convergence.   Consider  the  infinite  series 

where  n\  means  l«2-3---w  and  is  read  "factorial  n"  Dis- 
regarding for  the  moment  the  first  term,  compare  the  sum  of 
the  next  n  terms, 

°"n  =  1  +  t-£  +  7-ir-H f- 


1-2       L2-3   '  '    1-2-3...W 

with  the  corresponding  sum  of  the  geometric  series, 


=  2--3-<2. 

The  terms  of  <rn  after  the  first  two  are  less  than  those  of  Sn  and 
hence 

<rn<Sn<  2. 

Inserting  the  discarded  term  and  denoting  the  sum  of  the  first 
n  terms  of  (4)  by  sn  we  have : 

no  matter  how  large  n  be  taken.     That  is  to  say,  sn  is  a  varia- 
ble that  always  increases  as  n  increases,  but  that  never  attains 


246  CALCULUS 

so  large  a  value  as  3.  We  can  make  these  relations  clear  to 
the  eye  by  plotting  the  successive  values  of  n  as  points  on  a 
line. 

Sl  =  l  =  1 

s2  =  l  +  l  =2 

^=1+1+1  =2.5 

^1  +  1  +  ^  +  ^  =  2.667 

s5  =  2.708,       s6  =  2.717,       s7  =  2.718,       s8  =  2.718. 

Thus  we  see  that,  when  n  increases  by  1,  the  point  represent- 
ing sn  always  moves  to  the  right,  but  never  advances  so  far  as 
the  point  3.  Hence  sn  approaches  a  limit  e  which  is  not  greater 
than  3,  and  the  series  is  convergent.  To  judge  from  the  com- 
puted values  of  sn,  the  value  of  e  to  four  significant  figures  is 
2.718,  a  fact  that  will  be  established  later. 

The  reasoning  by  which  we  have  inferred  the  existence  of  a 
limit  in  the  above  example  is  of  prime  importance  in  the  theory 
of  infinite  series  as  well  as  in  other  branches  of  analysis.  AVe 
will  formulate  it  as  follows. 

Fundamental  Principle.  Ifsn  is  a  variable  which  (1)  always 
increases  (or  remains  unchanged)  when  n  increases : 

sn,^isn,  n'>n; 

but  which  (2)  never  exceeds  some  definite  fixed  number,  A: 

no  matter  what  value  n  has,  then  sn  approaches  a  limit,  U: 

lim  *„  =  U. 

M=  00, 

The  limit  U  is  not  greater  than  A :       U^  A. 

u         a 

Si  So     Sa 

ii r  i  in 

Fig.  76 


CONVERGENCE  OF  INFINITE   SERIES  247 

EXERCISE 

State  the  Principle  for  a  variable  which  is  always  decreasing, 
but  which  remains  greater  than  a  certain  fixed  quantity,  and 
draw  the  corresponding  diagram. 

By  means  of  the  foregoing  principle  we  can  state  a  simple 
test  for  the  convergence  of  an  infinite  series  of  positive  terms. 

Direct  Comparison  Test.     Let 

u0  +  ul-\-  ••• 

be  a  series  of  positive  terms  which  is  to  be  tested  for  conver- 
gence. If  a  seco?id  series  of  positive  terms  already  known  to  be 
convergent  : 

can  be  found  whose  terms  are  greater  than  or  at  most  equal  to  the 
corresponding  terms  of  the  series  to  be  tested : 

un<an, 

then  the  first  series  converges  and  its  value  does  not  exceed  the 
value  of  the  test-series. 

For  let  sn  =  u0  +  uY  + \-  un_l} 

Sn  =  a0+a1-\ ha„_i, 

lim  Sn  =  A. 

Then  since  Sn<A        and        sn^.Sn, 

it  follows  that  sn  <  A. 

Hence  sn  approaches  a  limit  U^  A,  q.  e.  d. 

It  is  frequently  convenient  in  studying  the  convergence  of  a 
series  to  discard  a  few  terms  at  the  beginning  and  to  consider 
the  new  series  thus  arising.  That  the  convergence  of  the 
latter  series  is  necessary  and  sufficient  for  the  convergence  of 
the  former  is  evident,  since 

*»  =  (tto  +  «M hVi)  +  («J hVi) 

=     ■  u  +  sn_m. 


248  CALCULUS 

Here  u  is  constant  and  sn  will  converge  toward  a  limit  if  sn_t 
does,  and  conversely. 

EXERCISES 

Prove  the  following  series  to  be  convergent. 

1.  1  +  -  +  -+-H . 

2.  r  +  r4  +  r9+rm+  •  •-,     0^r<l. 

3     -I  +  A  +  1+... 
3*    3!  +  5!+7!  +       ; 

1.2T2.3T3-4T 
Suggestion :  Write  sn  in  the  form  : 

^■(i-8+g-i)+-+e-B-iIjm 


n  +  l 


5    -S-+JL+JL  + 

5'    l-2+3.4  +  5.6  + 


6-    ^  +  3i  +  P+'"- 


4.  Divergent  Series.  If  a  series  is  to  converge,  then  evi- 
dently its  terms  must  approach  0  as  their  limit.  For  other- 
wise the  points  sn  could  not  cluster  about  a  single  point  as 
their  limit.  Hence  we  get  the  following  exceedingly  simple 
test  for  divergence.  It  holds  for  series  whose  terms  are 
positive  and  negative  at  pleasure. 

If  the  terms  of  a  series  do  not  approach  0  as  their  limit,  the 
series  diverges. 

*  It  can  be  shown  that  this  series  converges  when  p  >  1 ;  cf.  Infinite 
Series,  §  6. 


CONVERGENCE   OF   INFINITE   SERIES  249 

This  condition,  however,  is  only  sufficient,  not  necessary,  as 
the  following  example  shows : 

^2^3^4^ 

If  we  strike  in  anywhere  in  this  series  and  add  as  many  more 
terms  as  the  number  that  have  preceded : 

1  +-k+-  ■    * 


n  +  1     n-\-2  n  +  n 

we  get  a  sum  >^.  For  each  term  just  written  down  is 
>  l/2w,  and  there  are  n  of  them.  If,  then,  we  can  get  a  sum 
greater  than  \  out  of  the  series  as  often  as  we  like,  we  can  get 
a  sum  that  exceeds  a  billion,  or  any  other  number  you  choose 
to  name,  by  adding  a  sufficient  number  of  terms  together. 
Hence  the  series  diverges  in  spite  of  the  fact  that  its  terms  are 
growing  smaller  and  smaller.  This  series  is  known  as  the 
harmonic  series. 

A  further  test  for  divergence  corresponding  to  the  test  of 
§  3  for  convergence  is  as  follows. 

Direct  Comparison  Test.     Let 

«o  +  «H 

be  a  series  of  positive  terms  which  is  to  be  tested  for  divergence. 
If  a  second  series  of  positive  terms  already  known  to  be  divergent : 

a<>  +  aH 

can  be  found  whose  terms  are  less  than  or  at  most  equal  to  the 
corresponding  terms  of  the  series  to  be  tested  : 

then  that  series  diverges. 

The  proof  is  similar  to  that  of  the  test  of  §  3  for  conver- 
gence and  is  left  to  the  student  as  an  exercise. 


250  CALCULUS 

EXERCISES 
Prove  the  following  series  to  be  divergent. 

i.  i+Jt+A:+J=+..% 

V2      V3      V4 

5.  The  Test-Ratio  Test.  The  most  useful  test  for  the  conver- 
gence or  the  divergence  of  a  series  is  the  following,  which 
holds  regardless  of  whether  the  terms  are  positive  or  negative. 
It  makes  use  of  the  ratio  of  the  general  term  to  its  predecessor, 
^n+i/wn>  —  the  test-ratio,  as  we  shall  call  it. 

The  Test-Ratio  Test.     Let 

tto  + «i+  •'•■ 

be  an  infinite  series  and  let  the  limit  approached  by  its  test-ratio 


be  denoted 

byt: 

lim 

n=oo 

Un+l  _  ^ 

un 

Then  if 

l«!<i, 

the  series  converges; 

a 

l'l>i, 

"       "     diverges; 

« 

\t\  =  i, 

the  test  fails. 

We  shall  prove  the  theorem  in  this  paragraph,  so  far  as  it 
relates  to  convergence,  only  for  the  case  that  the  terms  are  all 
positive.     Then  £ J> 0  and  \t\=t. 

Suppose  t  <  1.  Let  y  be  chosen  between  t  and  1 :  t  <  y  <  1. 
Since  the  variable  un+l/un  approaches  t  as  its  limit,  the  points 
representing  this  variable  cluster  about  the  point  t  and  hence 


CONVERGENCE  OF  INFINITE   SERIES  251 

ultimately, —  i.e.  from  a  definite  value  of  n  on:  ft^m, —  lie 
to  the  left  of  the  point  y : 


^<y,  n>m. 

un 

t      y 

H 1 1 

1 

Fig.  77 


Now  give  to  n  successively  the  values  m,  m  + 1,  etc. : 


—  <y>  Wm+1<^roy; 


ft  =  m  +  l,  -=±*<y,  *W«<«Wiy<«*/; 


''TO+l 


n  =  m  +  2,  -=±*<y,  Wm+3<Wm+2y<Wmy3; 


u 


»»+2 


Hence  we  see  that  the  terms  of  the  given  series,  from  the 
term  um  on,  do  not  exceed  the  terms  of  the  convergent  geo- 
metric series 

wOT  +  wwy  +  wrny2+  ...f 

and  therefore  the  given  series  converges.* 

Secondly,  let  1 1 1  >  1,  the  terms  now  being  either  positive  or 
negative.     Then,  when  n  ^  m, 

K±lJ>l       or         K+1|>|«n|, 

I  Un  | 

i.e.  all  later  terms  are  numerically  greater  than  the  constant 
ftm,  and  so  they  do  not  approach  0  as  their  limit.  Hence  the 
series  diverges. 

*  The  student  should  notice  that  it  is  not  enough,  in  order  to  insure 
convergence,  that  the  test-ratio  remain  less  than  unity  when  n^m. 
Thus  for  the  harmonic  series  un+\/un  =  n/{n  +  1)<1  for  all  values  of  n, 
and  yet  the  series  diverges.  But  the  limit  of  the  test-ratio  is  not  less 
than  1.  What  is  needed  for  the  proof  is  that  the  test-ratio  should  ulti- 
mately become  and  remain  less  than  some  constant  quantity,  y,  itself  less 
than  1. 


252  CALCULUS 

Lastly,  if  1 1 1  =  1,  we  can  draw  no  inference  about  the  con- 
vergence of  the  series,  for  both  convergent  and  divergent  series 
may  have  the  limit  of  their  test-ratio  equal  to  unity.  Thus 
for  the  harmonic  series,  known  to  be  divergent : 

g»±?  =  — ?L^  =     1     ,  lim^l^l; 

n 
while  for  the  convergent  series  of  §  3,  Ex.  6  : 

^■^/-JL-V,         and        lim^±i  =  l. 

un     w  +  iy  »=»  un 


EXERCISES 

Test  the  following  series  for  convergence  or  divergence. 

1^34 

1.    2  +  f2  +  2~3  +  24+""  AUS'   Convergent 

1.2     1.2.3     1.2.3-4  A        ~. 

2     4- 4- -I .  Ans.    Divergent. 

1.1-2     L2.3  2^     3100     4100 

'    3     3-5     3-5-7  '     2        22       23 

4.    *+*  +  *+.;<,  6.    ^  +  -^  +  ^4-.... 

25  r  210     215  53     103     153 

For  what  values  of  x  are  the  following  series  convergent, 
for  what  values  divergent  ? 

7.  1  +  ^4^+.'..,  9.   1 +-  +  -  +  -+  .... 

8.  .T3  +  a5  +  a:7+....  10.    l-fz2+— +  —  +.... 

2!      3! 

6.   Alternating  Series.     Theorem.     Let  the  terms  of  an  in- 
finite series  be  alternately  positive  and  negative : 

u0  —  ux  -\-  a2  —  ••'. 


CONVERGENCE   OF  INFINITE   SERIES  253 

If  (1)  each  u  is  less  than  or  equal  to  its  predecessor :      un+1<un, 

and  (2)  lim  un  =  0, 

the  series  converges. 

For  example : 

1  _  1  4-  i  _  1  -I-  . . . 

Denote  as  usual  the  sum  of  the  first  n  terms  by  sn.  Then, 
when  n  is  even,  n  =  2m,  we  have  : 

S2>n  =  (««  —  «*l)  +  (U2  ~  U3)  H h  (Wom-2  —  MjJm-l)- 

Thus  s2/«  always  increases  or  remains  unchanged  when  m 
increases. 

If  n  is  odd,  n  =  2m  +  l, 

and  we  see  that  s2m+1  steadily  decreases  or  remains  unchanged 
when  m  increases. 

Furthermore,  s2m  does  not  exceed  the  fixed  value  8V     For 

S2m  =  ,S2»i+l  U2m  =  S2m+1  =  Sl  ■ 

Hence,  by  the  Fundamental  Principle  of  §  3,  s2l)l  approaches  a 
limit. 

In  like  manner  it  is  shown  that  s2m+1  is  never  less  than  s2. 
For 

S2m+l  =z  S2m  ~f"  U2m  ^  S2m  ^  S2  ■ 

Hence  s2m+l  also  approaches  a  limit. 

Finally,  these  limits  are  equal.     For,  since 

s2m+i  =  s2m  +«**,         lim  s2w+1  =  lim  s2m  +  lim  w2m, 

and,  by  hypothesis,  lim  un  =  0.  Hence  sn  approaches  a  limit 
when  n  becomes  infinite  passing  through  both  odd  and  even 
values,  and  the  series  converges,  q.  e.  d. 

It  is  easily  seen  that  the  error  made  by  breaking  an  alter- 
nating series  off  at  any  given  term  does  not  exceed  numeri- 
cally the  value  of  the  last  term  retained. 


254  CALCULUS 

7.  Series  of  Positive  and  Negative  Terms ;  General  Case.    Let 

<rm  =  v0  +  vl-] \-vm_x 

be  the  sum  of  the  first  m  positive  terms  of  the  series  (2), 

—  Tp  =  —  W0  —  W1 Wp_Y 

the  sum  of  the  first  p  negative  terms.      Then  sn  can,  by  a  suit- 
able choice  of  m  and  p,  be  written  in  the  form : # 

,        Sn=am  —  rp. 

For  example,  if  the  w-series  is 

W  +  W+-, 

the  ^-series  will  be 

and  the  —  w-series : 

__  i  _  i  _  i  __  ... 

When  n  =  oo,  m  and  p  will  in  general  both  increase  without 
limit,  and  two  cases  can  arise. 

Case  1.     Both  <rm  and  rp  approach  limits  : 

lim  arm  =  V,  lim  Tp  =  TF; 

m=co  p=oo 

i.e.  both  the  v-series  and  the  w-series  converge.     In  this  case 
the  ^series  converges, 

lim  sn  =  U,  and  U=  V-  W. 

n—co 

Case  2.  At  least  one  of  the  variables  <rm,  rp  diverges  when 
n  =  oo.  In  this  case  the  w-series  may  still  converge,  as  the 
above  example  shows.  But  if  one  of  the  auxiliary  series  con- 
verges and  the  other  diverges,  it  is  evident  that  sn  can  approach 
no  limit.     Example : 

l-r-hi-r2  +  i-r3  +  --,  0<r<l. 

Absolutely  Convergent  Series.  Let  us  form  the  series  of  the 
absolute  values  of  the  terms  of  the  ^-series : 

*  If,  for  a  given  value  of  w,  no  positive  terms  have  as  yet  appeared,  we 
will  understand  by  a0  the  value  0.     Similarly,  t0  =  0. 


CONVERGENCE  OF  INFINITE  SERIES  255 

KI  +  KI  +  -.  . 

Here  |  un\  will  be  a  certain  v  if  un  is  positive,  a  certain  w  if  un 
is  negative.     If  we  set 

s«=KI  +  lwil+  •••  +  lw»-i|> 

then  s'n  =  (Tm  +  V 

Hence  the  series  of  absolute  values  converges  if  both  the 

v-series  and  the  ly-series  converge. 

Conversely,  if  the  series  of  absolute  values  converges,  then 
both  the  v-series  and  the  w-series  converge  and  we  have  Case  1. 
For  both  of  the  latter  series  are  series  of  positive  terms,  and 
no  matter  how  many  terms  be  added  in  either  series,  the  sum 
cannot  exceed  the  value  U'  of  the  series  of  absolute  values. 
Hence  by  the  Principle  of  §  3  each  of  these  series  converges. 

Series  whose  absolute  value  series  converge  are  said  to  be 
absolutely  or  unconditionally  convergent;  other  convergent  series 
are  conditionally  convergent. 

We  can  now  complete  the  proof  of  the  theorem  of  §  5, 
namely,  for  the  case  that 

lim^±!=*,  |*|<1. 

«=*>   un 

Here  the  series  of  absolute  values  converges,  for 


and  hence        lim  \^n±ll  =  \t  I  <  1. 

...    \un\ 


Consequently  the  w-series  converges  absolutely. 
Example  1.     To  test  the  convergence  of  the  series 

x2  .  x3 

x 

2      3 

Here  ^»±i= —x,  lim£s±!=-^ 

un  n  +  1  n=w    un 

and  hence  the  series  converges  when  —  1  <  x  <  1  and  diverges 
outside  of  this  interval. 

Divergent  — 1 0 1  Divergent 

Convergen t 


256  CALCULUS 

At  the  extremities  of  the  interval  the  test  fails.  But  we  see 
directly  that  for  x  =  l  the  series  is  a  convergent  alternating 
series;  for  x  =  —  1,  the  negative  of  the  harmonic  series,  and 
hence  divergent. 

Example  2.     The  series 

^1-2  T  1-2.3  ^ 

has  for  its  general  term,  uk : 

_  n(w-l)  •••  (n  -fc-f  1)  „* 

If  n  is  a  positive  integer,  the  later  terms  are  all  0  and  the 
series  reduces  to  a  polynomial,  namely  the  binomial  expansion 
of  (1  +  x)H.  When  n  is  not  a  positive  integer,  the  value  of  the 
test-ratio  is 

1h±l  =  rLnlX)         and         \\mv^=-x. 

Uk  K  +  1  fc  =  Q0     Uk 

Hence  the  series  converges  when  —  1  <  x  <  1  and  diverges 
when  |  x  |  >  1.  For  the  determination  of  whether  the  series 
is  convergent  or  divergent  at  the  extremities  of  the  interval 
of  convergence  more  elaborate  tests  are  necessary. 

EXERCISES 

Eor  what  values  of  x  are  the  following  series  convergent  ? 
Indicate  the  interval  of  convergence  each  time  by  a  figure. 


/ 


1.   l  +  a:  +  2<ri  +  3ai3+—.  Ans.    —  1<z<1. 

Ans.    —  oo  <  x  <  oo,  i.e.  for  all  values  of  x. 

3.  tt_^  +  ^_^_j-....  Ans.    -l<a<L 

3      5      7 

4.  l_^  +  ^_^+.... 

2!      3!      4! 


CONVERGENCE  OF  INFINITE   SERIES  257 

5.   2.1a  +  3.2arJ  +  4.3arJ+— . 
v?   ,  a*5 

7.  »  +  - —  H — -  +  •*•• 

V3      V5 

8.  10  x  + 100  ^-f  1000a3 +  .... 

9.  a;  +  2»a?  +  48Ba!4  +  699a?+.... 
10.   l  +  a  +  2!a2  f  3!a3+.... 

12.  .+J+i  +  .... 

13.  lt^  +  ^|#+j-^|aC+.... 

,  Ice3   ,    L3x5  , 

15.    i  _________.... 

8.   Power  Series.     A  series  proceeding  according  to  mono- 
mials in  x  of  positive  and  steadily  increasing  degree : 

«o  +  «i»  +  a2x2  +  -.., 

is  called  a  power  series.  Such  a  series  may  converge  for  all 
values  of  x  or  for  no  value  of  x  except  0 ;  or  it  may  converge 
for  some  values  of  x  different  from  0  and  diverge  for  others. 
In  the  latter  case  the  interval  of  convergence  always  reaches 
out  to  equal  distances  on  each  side  of  the  point  x  =  0. 

This  latter  statement  is  easily  proven  for  such  power  series 
as  ordinarily  arise  in  practice.     If  we  assume,  namely,  that 


258  CALCULUS 

the  ratio  of  two  successive  coefficients,  an+1/anf  approaches  a 
limit : 

lim^±l  =  £, 

n  =  =o     an 

then  the  test-ratio  test  gives : 

lim  &+1  =  lim^s±i  x  =  Lx. 

n=oo     Un  n=oo     (Xn 

Hence  if  L  =  0,  the  series  converges  for  all  values  of  x ;  but 
if  L  ^=  0,  the  series  converges  when 

\Lx\<l,         i.e.         -|i|<aj<|i|, 

and  diverges  outside  this  interval. 

9.  Operations  with  Infinite  Series.  Since  the  value  of  an 
infinite  series  is  not  that  of  a  fixed  polynomial,  but  is  the  limit 
of  a  variable  polynomial,  we  cannot  expect  that  the  ordinary 
algebraic  processes  that  leave  the  value  of  a  polynomial  un- 
changed, such  as  rearranging  the  order  of  its  terms,  will  always 
leave  the  value  of  the  series  unchanged.  Nevertheless  it  can 
be  shown  that  the  terms  in  an  absolutely  convergent  series  can 
be  rearranged  at  pleasure  without  changing  the  value  of  the 
series.  Moreover,  any  two  convergent  series  can  be  added 
term  by  term : 

17=110  +  11!+    •••, 

V=v0  +  vl  H , 

ZT+V=:uQ  +  v0  +  ul  +  v1+  u2  H . 

And  two  absolutely  convergent  series  can  be  multiplied  to- 
gether like  polynomials : 

jJV=u0vo  +  uov1  +  ulvo+  w0^;24-^*1'u1^-^*2'yo^ • 

Hence,  in  particular,  for  power  series,  if 

f(x)  =  a0  +  axx  +  a2x2  +  •••, 

<£  (x)=  b0  +  bxx  +  b2x2  +  .-., 
then 

f(x)<f>(x)  =  a0b0  +  (a0b1  +  a1b0)x  +  (a0b2  +  aA  +  a2b0)x2+  .... 


CONVERGENCE  OF  INFINITE   SERIES  259 

The  resulting  series  thus  obtained  will  converge  at  least  for 
all  values  of  x  lying  within  the  smaller  of  the  two  intervals  of 
convergence  of  the  given  series. 

It  is  even  possible  to  divide  one  power  series  by  another  as 
if  they  were  both  polynomials.  We  shall  make  use  of  this 
property  in  the  next  chapter  when  we  come  to  develop  tan  x. 

An  especially  important  operation  with  power  series  is  that 
of  differentiating  or  integrating  the  series  term  by  term,  i.e.  as 
if  it  were  a  polynomial.  For  example,  take  the  geometric 
progression : 

—  =  l  +  «  +  ^  +  ^+  •-. 
1  —  x 

Differentiating  each  side  with  respect  to  x,  we  have 

(1  —  xy 

a  result  that  can  easily  be  verified  by  multiplying  the  first 
series  by  itself  as  explained  above. 

Again,  integrating  each  side  of  the  equation 

— —  =  l-x  +  x2-x*  +  ... 
1  +  x 

between  the  limits  0  and  h,  we  get,  since 
dx 


! 


% 


1  +  x  =  log(l  +  X) 


=  log(l  +  A), 


the  important  series : 

log(l  +  70  =  A-|  +  |--. 

By  means   of  this  series  and  others  immediately  deduced 
from  it  natural  and  denary  logarithms  are  computed. 
In  like  manner  we  get  from  the  series 

=  l-x2  +  3ti 


1  +  x* 
a  series  for  tan-1  ft: 


260  CALCULUS 


h 


1-f-z2  3      5 


By  means  of  this  series  the  value  of  ir  can  be  expeditiously 
computed  with  great  accuracy. 

It  is  of  value  for  the  student  at  this  stage,  before  proceeding 
to  the  further  study  of  series,  to  see  how  the  simpler  series  are 
actually  used  in  practice  as  a  means  of  computation.  He  is 
referred  for  a  treatment  of  this  subject  to  the  Infinite  Series, 
Chap.  II :  "  Series  as  a  Means  of  Computation,"  see  the  foot- 
note at  the  beginning  of  this  chapter. 

The  processes  with  infinite  series,  of  which  we  have  given  a 
brief  account  in  this  paragraph,  are  also  taken  up  and  estab- 
lished in  the  Infinite  /Series,  Chap.  IV :  "  Algebraic  Transforma- 
tions of  Series,"  and  Chap.  V :  "  Continuity,  Integration,  and 
Differentiation  of  Series."  In  the  latter  chapter  will  also  be 
found  a  proof  of  the  theorem  that  a  power  series  always  repre- 
sents a  continuous  function  throughout  its  whole  interval  of 
convergence. 

EXERCISES 

1.  If  ao  +  Oi  +  --- 

is  any  absolutely  convergent  series  and  p0>  Pi>**'   any  set  of 

numbers,  positive  or  negative,  that  merely  remain  finite  as  n 

increases:    \pn\<G,  where   G  is   a  constant,  show  that  the 

series 

«opo  +  «iPiH 

converges  absolutely. 

2.  Prove  that  the  series 

.    _     sin Sx  .  sin5aj 
SmX 3^  +  ^ 

converges  absolutely  for  all  values  of  x. 


CONVERGENCE   OF  INFINITE   SERIES  261 

3.   If  a0  -+-  «!  H and  6j  +  b2  H are  any  two  absolutely 

convergent  series,  the  series 

a0-|-a1cosic  +  a2cos2a;4-  ••• 

and  61sinic4-62sin2ajH 

converge  absolutely. 

^  4.    Show  that  the  series 

e~zcos  x  +  e~2a5cos  2x-\ 

converges  absolutely  for  all  positive  values  of  x. 

5.  What  can  you  say  about  the  convergence  of  the  series 

lH-rcos0+r2cos20+.--  ? 

6.  If  «o  +  «iH 

is  an  absolutely  convergent  series  and  if 

Wo  +  WxH 

is  a  series  such  that  un/a„  approaches  a  limit  when  n  =  oo , 
show  that  the  latter  series  converges  absolutely. 

7.  State  and  prove  an  analogous  theorem  for  divergent  series. 

8.  Show  that  the  series 

2x  2x     ,     2x     , 

7,-r  z o  +  K 9+  ••- 


1  -  aj*  '  4  —  **  '  9  —  a* 

converges  for  all  values  of  x  for  which  its  terms  all  have  a 
meaning. 

9.    Show  that  the  series 

a  a  a 

5  +  c     5  +  2c     6+3c+  '"' 

where  a  and  c  are  =£  0,  diverges. 
10.   Is  the  series 

convergent  or  divergent? 


CHAPTER  XIII 
TAYLOR'S  THEOREM 

1.  Maclaurin's  Series.  The  examples  to  which  the  student 
hasv  been  referred  in  the  preceding  paragraph  show  how  useful 
it  is  for  the  purposes  of  computation  to  be  able  to  represent  a 
function  by  means  of  a  series.  Such  a  representation  is  also 
important  as  an  aid  in  studying  properties  of  the  function. 
We  turD  now  to  a  general  method  for  representing  any  one 
of  a  large  class  of  functions  by  power  series,  — for  developing 
the  function  in  a  power  series,  to  use  the  ordinary  expres- 
sion. 

Suppose  that  it  is  possible  to  develop  a  function  in  a  power 

series : 

f(x)  =  c0  -f  cxx  +  c2x2  -f  • •• . 

What  values  will  the  coefficients  have?     If  we  set  #  =  0  we 
see  that 

/(0)  =  c0, 

and  thus  the  first  coefficient  c0  is  determined. 
To  get  the  next  coefficient,  differentiate: 

f'(x)  =  c1  +  2c2x  +  3c3x2-\-  •••, 

and  again  let  x  =  0 : 

/'(0)  =  Cl. 

Thus  q  is  found.     Proceeding  in  this  manner  we  obtain : 

f"(x)  =  2-lc2  +  3-2csx  +  4:-3cix2+  ..., 

/"(0)  =  2.1c2,  c2=«, 

262 


TAYLOR'S   THEOREM  263 

and  so  on ;  the  general  coefficient  having  the  value 

C"~     nl 

Hence  we  see  that,  if  f(x)  can  be  developed  in  powers  of  x,. 
the  series  will  have  the  form : 

(1)  /(a;)==/(0)+/'(0)x  +  ^|p^+.... 

This  series  is  known  as  Maclaurin's  Series. 
For  example,  let  y./^  _  ^ 

Here         /r(*)  =  e*,        f"tp)  =  f,     •••    /«(»)«*, 

and  /(0)=1,        /'(0)=1,        /"(0)  =  1,     etc. 

Hence  the  development  will  be  as  follows : 

(2)  e*  =  l+x  +  ±  +  t.+  .... 
This  series  converges^for  all  values  of  x. 

EXERCISES 

Assuming  that  the  function  can  be  developed  in  a  Mac- 
laurin's  Series,  obtain  the  following  developments. 

1.  sina  =  x--  +  --.... 

2.  «■•■!__+_-.... 

4.  (i+8).Bi.HM+"^|D^+»("-j1)(;-i>W.... 

Obtain  three  terms  in  each  of  the  following  developments. 

5.  tan#  =  #-f  Ja53  +  T?5«5+  •••• 

6.  secx  =  l+%x2  +  -%\x4 '■-. 

7.  c§iHX  =  l-f  a  +  ^o2— £a4+ •-•. 


264  CALCULUS 

2.  Taylor's  Series.  It  may,  however,  happen  that  no  devel- 
opment according  to  powers  of  x  is  possible.     Thus  if 

/(aj)=loga>, 

/(0)  =  —  oo.  But  a  power  series  represents  a  continuous  func- 
tion and  so  no  power  series  in  x  can  be  expected  to  represent 
log  x.  It  is  evident  generally  that,  whenever  the  function  or 
any  one  of  its  derivatives  becomes  discontinuous  for  x  =  0,  the 
function  cannot  be  developed  in  a  Maclaurin's  Series. 

A  power  series  is  most  useful  for  computation  if  the  values 
we  have  to  assign  to  its  argument  (i.e.  the  independent  vari- 
able) are  small.  Now  it  may  happen  that  we  know  the  value 
of  the  function  and  of  all  its  derivatives  at  a  single  point, 
x  =  x0,  or  at  least  can  easily  compute  them.  In  such  a  case 
we  can  find  the  value  of  the  function  at  points  x=x0  +  h  near 
by  if  we  develop  /(a?),  not  according  to  powers  of  x,  but 
according  to  powers  of  h.     Setting,  then, 

x  =  x0  +  h,  h  =  x  —  x0, 

we  shall  have,  if  a  development  be  possible : 

fix)  =f(x0  +  h)  =  c0  +Ci/i  +  c2h2  H . 

We  can  determine  the  coefficients  here  as  in  the  case  of  Mac- 
laurin's Series.     Thus,  setting  h  =  0,  we  find, 

f(x0)  =  c0. 

Differentiating  with  respect  to  h  and  remembering  that  x0  is 
a  constant,  we  obtain  : 

df(x)  r==df(x)dx  =  ,„.  = 
dh  dx    dh        K  J 

f'(x0  +  h)=:cl  +  2c2h  +  3c3h2+  -., 

/'f(aj0+7i)  =  2.1c2  +  3.2c3/i4-4.3c4^2H-  ••., 
/"(a0)=2.1c2,  c2=^f^ 


TAYLOR'S   THEOREM  265 

A  /(n)0<>) 

and  so  on :  cn  =J- — ^-yz 

n\ 

If,  then,  f(x)  can  be  developed  in  powers  of  h,  the  series  will 
have  the  form : 

(3)  fix,  +  h)  =/O0)  +/'(z0)  h+f-^£  tf  +  . ... 

When  h  is  replaced  by  x,  (3)  becomes  : 

(3')        /(»)  =/Oo)  +/,(%)(*-ab)  +-^f)(*-*„)2+  •- 

These  series  are  known  as  Taylor's  Series. 
For  example,  let 

/(a?)  =  logo?,  x0  =  l. 

Then 


™-1 

/'(i)  =  i; 

rwr>& 

/"(i)  =  -i; 

/"(-)=—, 

/"'(l)  =  2!; 

f™(x\-(     W+i<n- 

-1)! 

f<">m  =  r-n»+1('«- 

and  the  series  will  have  the  form : 

log(l+A)  =  A-|-  +  |— -. 

This  agrees  with  the  result  obtained  by  integration  in  the  pre- 
ceding chapter.  The  series  converges  for  values  of  h  numer- 
ically less  than  1. 

Maclaurin's  Series  is  a  special  casejrf  Taylor's  Series,  ob- 
tained by  letting  x0  =  0.  But  conversely,  Taylor's  Series  can 
be  obtained  from  Maclaurin's  by  replacing  x  by  h  as  above  and 
developing  f(x0  +  h)  in  a  Maclaurin's  Series. 


266  CALCULUS 


EXERCISES 

Assuming  that  the  function  can  be  developed  in  a  Taylor's 
Series,  obtain  the  following  developments. 

1.  ea+h=ea  +  eah  +  —Ji2+  .... 

21 

2.  sin  (x0  +  h)  =  sin  x0  -f  h  cos  x0  —  —  sin  x0  —  —-  cos  x0  +  •  • .. 

&  •  o ! 

V*       ;     V2L  2!^3!^      J 

4.    xn  =  (a  +  fc>"  =  a!'  +  j*a*-lft  +  ^^f^—  an  2h2 

^  1.2-3  a      fl  "^       * 

^6       7     2T   2  22!        2   3! 

6.  log*  =  log2+^_ife^  +  ^fc^)-3--. 

2         2       22  3       23 

Obtain  three  terms  in  the  development  of  each  of  the  fol- 
lowing functions. 

7.  logCl+z2),     x0  =  3. 

Ans.   2.303  +  .6  (x  -  3)  -  .08  (a  -  3)2  +  • .  -. 


8.  tanx,     x0  =  -'  10.     ,    x0=  —  1. 

4  ic 

9.  log(ex  +  e"x),  a0  =  0.  11.    10x,     x0  =  0. 

3.  Proof  of  Taylor's  Theorem.  Let  the  function  f(x)  be 
continuous  throughout  the  interval  a  <  x  <  b  and  let  it  have 
continuous  derivatives  of  all  orders  throughout  this  interval. 
Let  x0  be  an  arbitrary  point  of  the  interval,  which,  once 
chosen,  shall  be  held  fast,  and  let  x0  -f  h  be  any  second  point 
of  the  interval.  We  will  see  if  we  can  approximate  to  the 
value  of  the  function  by  means  of  the  first  n  + 1  terms  of  the 
corresponding  Taylor's  Series : 


TAYLOR'S  THEOREM  267 

(4)  /  (3%  +  h)  =/O0)  +  f(x0)  *+'-+  "CifiSJ  *"  +  5, 


n 


where  R  denotes  the  error,  i.e.  the  difference  between  the 
value  of  the  function  and  the  value  of  the  approximation. 
In  order  to  see  how  good  this  approximation  is,  we  must  have 
an  expression  for  R  that  will  throw  light  on  the  numerical 
value  of  this  quantity.  Such  an  expression  can  be  found  as 
follows. 

Let  us  write  R  in  the  form  : 

7»«+l  7,n+l 

R  =  — P,         i.e.  let        P=R+ 


0  +  1)!  0  +  1)! 

Then  (4)  becomes,  on  transposing  terms  : 

(5)  f(x,  +  h)  -/(*„)   -  hf  (*) £/">(*.)  -  j^-P=0. 

n\  (ti+1)! 

We  now  proceed  to  form  arbitrarily  the  following  function 

of  z : 

*  (s)  =/(X)  _/<i)  -  (X -  z)f  (z)  -  ^ff^f"  (,)--.. 

Here  X  =  x0-j-h,  and  X  and  P  are  constants.  This  function 
satisfies  all  the  conditions  of  Rolle's  Theorem  in  the  interval 
®oSz^X.  For  <f>(X)  is  obviously  =0,  and  if  we  compare 
<f>  (a?0)  with  the  left-hand  side  of  (5),  we  see  that  <f>  (x0)  vanishes, 
too.  Hence  the  derivative  of  <f>  (z)  must  vanish  at  some  point 
within  the  interval.  Now,  on  computing  the  derivative  we 
find  that  the  terms  cancel  each  other  to  a  large  extent :  * 

<*>'  (?)  =  ~f  (*)  +  /'  (•)  -  (X-  z)f"  (z)  +  (X  -  z)f»  (z)  - 

(-X'  —  g)nf(n+l)/g\  _|_  (X—Z)np^ 

so  that  there  remain  finally  only  two  terms : 

*  The  student  is  requested  to  write  out  the  terms  in  this  differentiation 
for  n  =  1,  2,  and  3. 


268  CALCULUS 

n\  n\ 

Consequently  the  conclusion  of  Rolle's  Theorem: 

4>'(Z)  =  0,        x0<Z<X      or       Z^xo+dh,     0<0<1, 
leads  to  the  result, 

(6)  P  =  /<-+»fo  +  M),  E^-^-f^ixo  +  Oh). 

{n  + 1) ! 

Thus  we  obtain  one  of  the  most  important  theorems  of  the 
Calculus,  Taylor's  Theorem  with  the  Remainder :  * 

(7)  f(x0  +  h)=f(x0)  +/(*o)/*+^^2+--  +^A" 

~_'  ft  • 

If  we  set  w  =  0,  thus  stopping  with  the  second  term,  we  get 
the  Law  of  the  Mean : 

f(x0  +  h)  =f(x0)  +  hf  (x0  +  Oh). 
If  n  =  1,  we  have : 

/O0  +  70  =/(*0)  +  hf  (x0)  +  |^/''(a%  +  Oh). 

If  we  allow  n  to  increase  without  limit,  the  first  n  + 1  terms 
of  (7)  become  an  infinite  series,  the  Taylor's  Series  corre- 
sponding to  the  function  f(x).  In  order  that  this  series  should 
converge  and  represent  the  function  it  is  necessary  and  sufficient 
that 

(8)  limJK  =  0. 

n=oo 

When  the  condition  (8)  is  satisfied,  we  say  that  the  function 
can  be  developed  or  expanded  by  Taylor's  Theorem  about  the 
point  x  =  x0. 

*  In  the  foregoing  proof  we  have  made  no  use  of  that  part  of  the 
assumption  regarding  f(x)  which  relates  to  derivatives  of  higher  order 
than  n  +  1,  and  consequently  our  theorem  is  somewhat  more  general  than 
would  appear  in  the  text. 


TAYLOR'S  THEOREM  269 

4.  A  Second  Form  for  the  Remainder.     A  form  of  the  re- 
mainder which  is  obtained  by  setting 

R  =  hP, 

and  proceeding  as  in  §  3,  is  sometimes  useful.     Thus  we  have 

/(3b  +  h)  -f(x0)  -  hf>  (x0) *L-f»\ (x0)  -hP=0, 

ni 

and  we  form  the  function  of  z, 

$  (z)  =/(X)  -/(*)  -  (X  -  z)f  (z)  -  (X^f"(z)  - 

where  X—x0-{-h.  This  function  satisfies  the  conditions  of 
Rolle's  Theorem  in  the  interval  x0  fj  z  <  X,  and  so  its  deriva- 
tive, *'  (z)  -  -  (X~Z)V(W+1)  (*)  +  P, 

must  vanish  at  some  point  Z  =  x0  +  0h  within  the  interval. 

Hence, 

(9)  B  =  0--°yhn+1f(n+»  (^  +  ehy 

5.  Development  of  ex,  sin  x,  cos  x.     The  function  e*  can  be 
developed  by  Taylor's  Theorem  about  the  point  x0  =  0.     Here 

/(V)=6*,  /'<*)»«",       •     •     •      /»(»)-*, 

/(0)  =  1,         /'(0)  =  1,      •   •  •    /™(0)  =  1, 
and  the  remainder  R  as  given  by  (6)  has  the  form : 

fcn+l 

i£  =  — e°\ 

(n  +  1)! 

If  h<0,         e0h<l,         and         R< 


\h\»+1 
0  +  1)! 


If  h>0,         e0h<e\        and         R<     7*"+1     e\ 

(n  +  1)! 

Now  lim     h*+1     =  0. 


For  we  can  write 


(«+i) 


270  CALCULUS 

hn+1     =  h  #  h  t  h  m  h         h 

(w  +  1)!      1  '  2  '  3  '         '  n  '  n  +  1 

No  matter  how  large  h  may  be  numerically,  since  it  is  fixed 
and  n  is  variable,  these  factors  ultimately  become  small,  and 
hence  from  a  definite  point  n  =  m  on 

1 — L<-,  n>m. 

n       2 

If  we  denote,  then,  the  product  of  the  first  m  factors,  taken 
numerically,  by  C,  and  replace  each  of  the  subsequent  factors 
by  !-,  we  shall  have : 

7,71+1         I  /1\n-m+l 


(n  + 1)  !  J  \2, 

The  limit  of  this  last  expression  is  0  when  n  =  <x>,  and  conse- 
quently *  lim  hn+1/  (n  + 1)  !  =  0. 

We  have,  then,  lim  R  =  0  and  hence,  replacing  h  by  x : 

n=oo 

(io)  ,-i+.+j£+!;+.... 

The  series  converges  and  represents  the  function  for  all  values 
of  x. 

To  develop  sin  x  we  observe  that 

f(x)=smx,  ,/(0)=0, 

/' (a^cosz,  /(0)  =  1, 

f"(x)  =  -smx,  /"(0)=0, 

/>"(X)=-C08X,  /'"(0)=-l, 

and  from  this  point  on  these  values  repeat  themselves. 

It  is  not  difficult  to  get  a  general  expression  for  the  n-th. 
derivative,  namely : 

*  We  might  have  given  a  short  proof  of  this  relation  by  observing  that 
fen+1  /  (n  +  1)  1  is  the  general  term  of  a  convergent  series : 

1  +  7*  +  —  +£-+.... 
21      3! 


TAYLOR'S   THEOREM  271 


/(«>(*)  =  sin^  +  ^\ 


This  formula  obviously  holds  for  n  =  1,  2,  3,  4,  and  from  that 
point  on  the  right-hand  member  repeats  itself,  as  it  should. 
Thus  we  find : 


R=     hn+]     Sm(  Oh 
0  +  1)!       V 


+ 


W7r\ 

2/ 


The  second  factor  is  never  greater  than  1  numerically,  and 
the  first  factor,  as  we  have  just  seen,  approaches  0  as  its  limit. 
Hence  lim  R  =  0  and  we  have,  on  replacing  h  by  x : 

n=xo 

(11)  sino^-^  +  fl-.... 

In  a  similar  manner  it  is  shown  that 

(12)  cosa  =  l-^  +  — . 

V     J  2r  4! 


EXERCISES 

1.  Compute  the  value  of  e06  (cf.  Chap.  IV,  §  7)  to  six  signifi- 
cant figures. 

2.  Show  that  ex  can  be  developed  by  Taylor's  Theorem  about 
any  point  x0. 

3.  Obtain  a  general  expression  for  the  n-th  derivative  of  cos  x 
and  hence  prove  the  development  (12). 

4.  Show  that  sin  x  and  cos  x  can  be  developed  by  Taylor's 
Theorem  about  any  point  x0. 

5.  Remembering  that  1°  is  equal  to  tt/180  radians,  compute 
sin  1°  correct  to  six  significant  figures.  By  about  what  percent- 
age of  either  does  sin  1°  differ  from  its  arc  in  the  unit  circle  ? 


272  CALCULUS 

6.   The  Binomial  Theorem.     Let 

f(x)=x», 

where  n  is  any  constant,  integral,  fractional,  or  incommensur- 
able, positive  or  negative ;  and  let  xQ  =  1. 
Then  /(1)  =  land 

f(x)  =  nx*-\  /'(l)  =  n, 

/' '  (a?)  =  n  (n  -  1)  xn~\        f"  (1)  =  n  (n  -  1), 


fW(x)  =  n(n  -  1)  ...  (n  -  k  +  l)xn~k, 

f»(l)  =  n(n-l).>.(n-k  +  l). 

For  the  remainder  R  it  is  better  here  to  employ  the  second 
form,  (9).     Thus 

B  =  0--°yhk+1 .  »(n - 1)  ...  (n  -  k)(l  +  ^)"~*-1 
1   k ! 

The  last  factor  remains  finite,  whatever  the  value  of  0,  pro- 
vided |  h  |  <  1.     For,  since   0  <  6  <  1, 

i-|ft|<i  +  M<i  +  IH 

and  by  Chap.  II,  §  8 : 

(l+tf^n-l  <(!   +  !/*  I)-!,  n>lj 

(i  +  ehy-*<(i-\h\y-\      n<i. 

The  next  to  the  last  factor  is  always  positive  and  less  than 
unity,  since  k  >  0  and 

D<i-=ii<l. 
1  +  fl* 

Finally,  the  remaining  expression  is  the  general  term  of  a 
series  already  shown  to  be  convergent,  namely  (cf .  Chap.  XII, 
§7): 


TAYLOR'S  THEOREM  273 


i  +  w&+a^|ay+»(»--1X*-8?»+...,    _!<*<!, 

and  hence  it  approaches  0  as  its  limit.     It  follows,  then,  that 
R  approaches  0  and  we  have  on  replacing  h  by  x : 

(13)      (l  +  xr  =  l  +  nx  +  n-(f^J  +  n(n-yn3-2'>«*  +  :... 

This  is  the  Binomial  Theorem  for  negative  and  fractional 
exponents.  When  n  is  0  or  a  positive  integer,  the  series 
breaks  off  of  itself  with  a  finite  number  of  terms  and  we  have 
a  polynomial,  namely :  (1  +  x)n.  In  all  other  cases  the  series 
converges  when  x  is  numerically  less  than  1  and  represents  the 
function  (1  -f-  x)n ;  and  it  diverges  when  x  is  numerically  greater 
than  1. 

The    following     developments     obtained    from 
especially  useful. 


■VI -x9 


1  *  L3 


(15)  yi-^  =  i-^-^-^*« 


EXERCISES 

1.    Show  that,  when  |  a  |  >  |  b  | : 

1  •  2i  1  •  £  >  o 

^2.    Compute  V3  correct  to  seven  significant  figures  by  means 
of  the  series  (13). 

Suggestion:     Eegin  by 'writing  3  =(f)2(ff).     Here  £  is  one  of  the 
convergents  in  the  development  of  V3  by  continued  fractions. 

3.  Compute  V30  to  five  significant  figures. 

4.  Obtain  from  (13)  the  development : 

l_^_2z  +  3^-4*3+-". 
(1  +  x)2 


274 


CALCULUS 


5.   Obtain  the  development : 

log(l  +  A)  =  A-|  +  |-- 
by  the  method  of  this  paragraph. 

7.  Development  of  sin-1  a?.  We  can  now  obtain  the  develop- 
ment of  sin-1  a;  in  a  manner  similar  to  that  employed  for  tan-1  x. 
Integrating  each  side  of  (14)  gives : 


'    ,  lh*  ,  l-.SVv 


r   dx 

J  vr= 


The  value  of  the  left-hand  side  is  sin-1  ft. 
h  by  x,  we  have  : 

1  a3  ,  1  •  3  ar5 


Hence,  replacing 


(16) 


sin  *o? 


X  +  23+f^L  5  + 


The  series  converges  and  represents  the  function  when  |  x  |  <  1. 

8.   Development  of  tan  x.    We  have : 

,  sin  a;      x  —  4-a^4- -A-naP—  ••• 

tan  x  = = o      \     , • 

cos  a?      1  —  -g-ar-f  ^ar  —  ... 

Now  it  can  be  shown  that  one  power  series  can  be  divided  by 
another  just  as  if  both  were  polynomials,  the  resulting  series 
converging  throughout  a  certain  interval,  cf.  Infinite  Series, 
§  36.     Hence 

1  ^2    l       1    /»4  .  _  _  <\ 
1' 


A*+A 


)x 


a;  —  lx?  +  J-  ar* 


^--^  + 
*  ^  + 


"3 

i*3 


We  can  obtain  in  this  way  as  many  terms  in  the  development 
of  tana;  as  we  wish,  although  the  law  of  the  series  does  not 
become  obvions. 
(17)  tdLnx  =  x  +  ^x3  +  -&x5-i 


TAYLOR'S  THEOREM  275 

9.  Applications.  We  shall  consider  here  only  two  or  three 
applications  of  Taylor's  Theorem,  referring  the  student  for 
further  applications  to  the  Infinite  Series,  Chaps.  II,  III,  and 
IV. 

(1)  Test  for  Maxima,  Minima,  and  Points  of  Inflection.  We 
can  now  state  wider  sufficient  conditions  for  maxima,  minima, 
and  points  of  inflection  than  those  given  in  Chap.  III. 

Suppose  that  the  function  f(x),  together  with  its  first  n  de- 
rivatives, is  continuous  in  the  neighborhood  of  the  point  x  =  x0 
and  that 

/'(*o)  =  0,         /"(z0)  =  0,      .   .   .     /(.-i) (a;0)=0, 

but  that  fn\x0)=t=0. 

Then  we  shall  have,  by  Taylor's  Theorem  with  the  Remainder, 

Formula  (7) : 

(18)  /  (a*  +  h)  -f(x0)  =  h«f^  (xo  +  Oh)/n !. 

If  n  is  even,  hn  will  be  positive  on  both  sides  of  the  point 
h  =  0,  x  =  x0 ;  and  since  f(n)  (x)  is  continuous,  it  will  preserve 
the  sign  it  has  at  x0  throughout  a  certain  interval  about  this 
point : 

x0  —  a < a; <#0 -f  a,  —  a<h  <a. 

Hence  the  right-hand  side  of  (18)  is  positive,  or  else  it  is  nega- 
tive, when  0  <  |  h  |  <  a  and  thus  we  are  led  to  the  following 

Test  for  a  Maximum  or  a  Minimum.     If 

/'(z0)=0,      f"(x0)  =  0,     •  •   •    /^-i>(z0)  =  0,      f^(x0)^0, 
the  function  f(x)  will  have 

a  maximum    at  x  =  x0   if  /(2m)  (x0)  <  0 ; 

a  minimum      "       "        "  /(2m)  (x0)  >  0. 

If,  on  the  other  hand,  n  is  odd,  the  right-hand  side  of  (18) 
will  change  sign  with  h  and  we  shall  have  a  point  of  inflection 
parallel  to  the  axis  of  x.  More  generally,  since  the  condition 
for  a  point  of  inflection,  be  it  parallel  to  the  axis  of  x  or  not,  is 
that  tan  t =/'(#)  be  at  a  maximum  or  a  minimum,  we  deduce 


276  CALCULUS 

from  the  test  just  obtained,  applied,  not  to/(#),  but  to /'(#)  = 
tanr,  the  following 

Test  for  a  Point  of  Inflection.   If 

/"O*o)=0,      f"'(x0)=0,     .  .  .    /»(ai)a0,      /(2OT+1)^o)^0, 

the  curve  y=f(x)  has  a  point  of  infection  in  the  point  (xQ,  yQ). 

(2)  Order  of  Contact  of  Two  Curves.  Let  two  curves,  Cl  and 
(72,  be  tangent  to  each  other  at  an  ordinary  point  P  of  either 
curve,  and  draw  their  common  tangent  PT.  At  a  point  M  of 
PT  infinitely  near  to  P  (by  this  is  meant  that  M  is  taken  con- 
veniently near  to  P  and  is  later  going  to  be  made  to  approach 
P  as  its  limit)  erect  a  perpendicular  cutting  CY  in  Px  and  C2  in 
P2.  PM  and  the  arcs  PPj ,  PP2  are  obviously  all  infinitesimals 
of  the  same  order.  It  will  be  convenient  to  take  PM  as  the 
principal  infinitesimal.  Denote  by  n  the  order  of  the  infini- 
tesimal PiP2.  Then  the  curves  Cx  and  C2  are  said  to  have 
contact  of  the  n  —  lst  order. 

For  example,  the  parabola 

c»  ft:  y  =  x> 

has  contact  of  the  first  order  with  its  tan- 


gent at  its  vertex : 

02:  2/  =  0. 

But  the  curve  y  —x^  has  contact  of  the  second  order  with  its 
tangent  at  the  origin ;  this  point  being  a  point  of  inflection  for 
the  latter  curve.     And  the  curves 

y  =  xsf  y  =  Xs  —  x* 

have  contact  of  the  third  order. 

Since  we  can  always  transform  our  coordinate  axes  so  that 
the  tangent  PT  will  be  parallel  to  the  axis  of  x  —  such  a  trans- 
formation evidently  has  no  influence  on  the  order  of  contact  of 
the  curves  —  we  may  without  loss  of  generality  assume  the 
equations  of  the  curves  in  the  form  * 

C,:  y  =/(*), 

02:  y  =  <j>(x), 


TAYLOR'S  THEOREM  277 

where      y0=/(«b)  =<£  0»o)        and     f'(x0)=0,       <f>'(x0)  =  0. 
Hence,  by  Taylor's  Theorem  with  the  Remainder,  (7)  : 

ci:        y-y.=^/"K)+-  +^fl")(*o+M), 
On         y-y^  ^<t>"  (^)+-  +  ^n)(^  +  e'h). 

The  infinitesimal  PXP2  on  which  the  order  of  contact  of 
these  curves  depends  is  numerically  equal  to  the  difference 
between  the  ordinate  y  of  Cx  and  the  ordinate  y  of  C2,  ie.  to 

(i9)  fr/'^o)-<n^)]+-  •  • 

+  j?f*»  (*o  +  Oh)  -  +<»>  (a*  +  M)~L 

Now  the  curvature  of  these  curves  at  the  point  (xQ,  y0)  is, 
since  /'  (x0)  =  0  and  </>'  (x0)  =  0 : 

*i=|/"te>)  |,  «s=|+"0*>)|. 

Hence  the  curves  will  have  contact  of  the  first  order  if  theyl 
have  different  curvatures  at  P,  or  if  they  have  the  same  curva- 
ture (#=()),  one  curve  being  concave  upward  and  the  other  con- 
cave downward.  But  if  they  have  the  same  curvature  and  (in 
case  the  curvature  of  both  is  =#=  0)  if  they  both  present  their 
concave  side  in  the  same  direction,  then  they  will  have  contact 
of  at  least  the  second  order.  Thus  at  an  ordinary  point  a 
curve  has  contact  of  the  first  order  with  its  tangent. 

In  particular,  let  C2  be  the  osculating  circle  of  Cx  at  P. 
Then  C2  has  the  same  curvature  as  Cx  and  is  concave  toward 
the  same  side  of  the  tangent.  Hence  it  has  in  general  contact 
of  the  second  order  with  d ;  but  at  special  points  it  may  have 
contact  of  higher  order. 

At  an  ordinary  point  of  inflection  the  tangent  line  has  con- 
tact of  the  second  order  with  the  curve.  For  here,  if  we  take 
C2  as  the  tangent  line,  <f>  (x)  =  0  for  all  values  of  x,  and  hence 
the  derivatives  <£"(a\,),  <t>'"(xo),  etc.  all  vanish.  On  the  other 
hand,  /"  (x0)  =  0,  /'"  (x0)  =£  0.     Consequently  (19)  becomes 


278  CALCULUS 

||/»'(ai  +  «)  and  lim  ££  =  $/»' (**,)*<). 

(3)   Evaluation  of  the  Limits  -,  oo  —  oo,  etc.     The  limit  of 
the  fraction 

lim^M, 
x=«  jF(a!) 

when  /(a)  =  0  and  F(a)  =0,  can  be  obtained  without  the  labor 
of  differentiating  whenever  the  numerator  and  the  denomina- 
tor can  be  expressed   as  power  series   in  terms  of  x  —  a  =  h. 

For  example,  to  find 

v     x  -  sin  x 

lim 

x±ox—  tana; 

By  the  aid  of  the  series  for  sin  x  and  tan  x,  we  have 

x  —  sin  x  _      ^  x3  +  higher  powers  of  x 
x  —  tan  x      —  ^  x3  +  higher  powers  of  x 

Hence,  cancelling  x3  from  the  numerator  and  the  denominator, 
we  see  that  the  value  of  the  limit  is  —  |. 

The  method  of  series  is  often  of  service  in  evaluating  the 
limit  oo  —  oo  .     For  example,  to  find 

lim  ( Vl  +  v?  —  x). 
Here  we  can  take  out  #  as  a  factor : 


(aK-> 


and  then  express  the  radical,  since  x  >  1,  as  a  series  in  1/x  by 
means  of  the  Binomial  Theorem : 


v 


.1-1+1.  1+3.I+ 


Hence  x(^l  +  J-l)-|  •  £  +  f  •  J  + 


TAYLOR'S   THEOREM  279 

When  x  =  <x>,  the  terms  of  this  power  series  in  1  /x  approach  0 
as  their  limit,  and  since  a  power  series  represents  a  continuous 
function,  the  value  of  the  limit  in  question  is  seen  to  be  0. 

EXERCISES 

1.  Show  that  the  function 

y  =  2  cos  x  +  x  sin  x 
has  a  maximum  when  x  —  0. 

2.  Have  the  following  functions  maxima,  minima,  or  points 
of  inflection  when  x  =  0  ? 

(a)  5  sin  x  —  4  sin  2  x  4-  sin  3  #. 
(6)  2ar5-3e*  +  6sina  +  -x. 

(c)  locosx  —  6  cos  2  a;  +  cos  3  a;. 

3.  Determine  all  the  maxima,  minima,  and  points  of  inflec- 
tion of  the  function 

y  —  \x  —  \  sin  x  +  y1^  sin  2x, 

and  hence  plot  the  graph. 

£..   Show  that  the  curve  y  —  cos  x  has  contact  of  the  fifth 
order  at  the  point  (0,  1)  with  the  curve 

y  =  l-$x?  +  ^x\ 

5.  Show  that  the  curve  y  =  sinx  has  contact  of  the  sixth 
order  at  the  origin  with  the  curve 

y  =  x-lx3  +  T%-ux5. 

6.  Determine  the  parabola 

y  =  a  +  bx-\-cx2 

so  that  it  shall  have  contact  of  the  second  order  with  the  curve 
y  =  ex,  when  x  =  0. 

7.  The  same  when  x  =  1.  Ans.  y  =  %e  +  ^ex2. 


280  CALCULUS 

8.  Show  that,  when  the  function  f(x)  is  represented  by  a 
Taylor's  Series,  the  n-th.  approximation  curve : 

y  =  sn(x)=f(x0)  +f(xo)  (x-x0)  +  ...  +f^^ (x-x0y, 

has  contact  of  at  least  the  n-th.  order  with  the  curve  y  =f(x) 
at  the  point  (x0,  y0).  When  will  it  have  contact  of  higher 
order  ? 

9.  Show  that  the  curve 

ax  +  P 

y      yx  +  8 

can  in  general  be  so  determined  as  to  have  contact  of  the 
second  order  with  the  curve  y=f(x)  at  the  point  (x0,  y0).     For 
simplicity,  assume  x0  =  0  and  y0  =  0. 
What  cases  are  exceptions  ? 

10.  Show  that 

o 
i 

(M      f^!-dx  =  ~ L-  +  -JL (a>0). 

W   J  1  +  x*  a      a  +  b^a  +  2b         '    V  ; 


(0j 


e~**dx  —  x  — -  -f- 


3       5   21       73! 

11.   Evaluate  to  three  significant  figures 


IT 

! 


since  , 
dx. 


Evaluate  the  following  limits : 
12.   limfcota; )•   Ans.  0.    13.  lim(Vl  +x  —  x).    Ans.  — oo. 

as  =  0   \^  XJ  *  =  oo 


TAYLOR'S   THEOREM  281 


i  1 

14.   limfcY.        Ans.   1.    16.   lim  fcY.  Ans.   0. 

x=Q  \     X     J  *  =  0    \     X     J 


15.   limf-1— ^)    •    Ans.  -^.    17.    lim(cosx)-.  1 

*±o\   x    J  ye  x^o  ^ng>    J_# 

Ve 

18.  Show  that,  when  two  curves  have  contact  of  even  order, 
they  cross  each  other ;  when  they  have  contact  of  odd  order, 
they  do  not  cross. 

19.  If  f(x)<<j>(x) 

is  .^/(s)  <!_*(*)      ? 

dx  dx 

20.  If  -^rJ->-^-L 

dx  dx 

and  f(x0)  =  tj>(x0)i 


is 


f(x0  +  h)^<j>(x0  +  h),         h>0     ? 


21.  Show  that  sin  a  —  a 

is  an  iufinitesimal  of  the  third  order,  referred  to  a  as  principal 
infinitesimal. 

_a2 

22.  Determine  the  order  of  the  infinitesimal  cos  a  —  e  2* 

23.  Show  that  the  equation 

<j>  sin  cf>  =  1 
has  one  and  only  one  root  lying  between  0  and  7r/2. 


r 


CHAPTER  XIV 
PARTIAL  DIFFERENTIATION 

1.  Functions  of  Several  Variables.    Limits  and  Continuity. 

We  shall  consider  in  this  chapter  functions  that  depend  on 
more  than  one  variable.  Thus  the  area  z  of  a  rectangle  is  the 
product  of  its  two  sides,  x  and  y : 

z  =  xy; 

and  the  volume  u  of  a  rectangular  parallelopiped  is  the  product 
of  its  three  edges  x,  y,  and  z : 

u  —  xyz. 

If  the  number  of  independent  variables  is  two,  we  can  rep- 
resent the  function 

(1)  *=/(*,  V) 

geometrically  as  a  surface. 

Such  a  function  is  said  to  be  continuous  at  the  point  (x0,  y0,  z0) 
if  a  small  change  in  the  values  of  x  and  y  gives  rise  only  to  a 
small  change  in  the  value  of  the  function.  And  the  function 
is  said  to  approach  a  limit,  z0,  when  the  point  (x,  y)  approaches 
(x0,  y0),  if  the  point  (x,  y,  z)  of  the  surface  (1)  approaches  a 
limiting  point  (x0,  y0,  zQ)  in  space,  no  matter  how  the  point 
(x,  y)  in  the  plane  may  approach  the  point  (a?0,  y0)  as  its 
limit. 

To  formulate  this  latter  definition  in  a  more  precise  manner 
and  at  the  same  time  in  a  way  that  is  applicable  to  functions 
of  more  than  two  variables,  let  c  be  an  arbitrarily  small  positive 
quantity.     If  a  positive  8  can  be  found  such  that 

282 


PARTIAL   DIFFERENTIATION 


283 


\f(x,y)-z0\<e 

for  all  points  (*,  y), —  except,  of  course,  (x0,  y0),  —  which  lie  in 
the  neighborhood  of  (sc0,  yQ) : 

x-x0\<$,  \y  —  yo\<&, 

then/(sc,  y)  i^ said  to  approach  z0  as  its  limit,  and  we  write: 

lim    f(x,y)  =  z0. 
■***  y-y0 

This  conception  once  being  made  precise,  we  can  now  render 

the  former  one  accurate  by  saying:  f(x,y)  is  continuous  at  the 

point  (xq,  y0)  if 

lim    f(x,y)=f(x0,y0). 

*=*0,  y=y0 

2.  Formulas  of  Solid  Analytic  Geometry.  In  what  follows 
we  shall  need  only  the  simplest  formulas  of  solid  analytic 
geometry,  and  we  set  them  down  here,  referring  the  student 
for  the  proofs  to  any  of  the  current  texts.  * 

Direction  Cosines.  If  a,  /?,  y  denote  the  angles  that  a  line 
makes  respectively  with  the  axes  of  x,  y,  and  z,  its  direction 
cosines  satisfy  the  relation : 

(2)  cos2  a  +  cos2  /?  +  cos2  y  =  1. 

The  angle  6  between  two  lines  is 
given  by  the  equation : 

(3)  cos  6  = 
cos  a  cos  a'  +  cos  (3  cos  ft'  +  cos  y  cos  y'. 

If  I,  m,  n  and  V,  ra',  n'  are  the  direc- 
tion cosines  of  two  lines,  or  quantities 
proportional  to  them  : 

I  =  p  cos  a,       m=p  cos  ft       n  =  p  cos  y ; 

then  the  necessary  and  sufficient  condition  that  the  lines  be 
perpendicular  to  each  other  is  that 


etc., 


*  Cf .  for  example  Bailey  and  Woods,  Analytic  Geometry,  p.  273  et  seq. 


284 


CALCULUS 


(4)  IV  +  mm1  +  nn'  =  0. 
The  condition  for  their  being  parallel  is  that 

(5)  l:l'  =  m:m'  =  n:  n'. 
Distance  Between  Two  Points : 

(6) 


V(aj!  -  x0)2  +  (yx  -  2/0)2  +  (zx  -  z0)2. 

Equation  of  Sphere.     Let  the  centre  be  at  (a,  b,  c)  and  the 
radius  be  r : 
(7)  (a;-a)8+(y-6)2+(2-c)2  =  r8. 

77ie  Plane.  Let  OP  be  the  perpendicular  dropped  from  the 
origin  on  the  plane,  let  OP=p,  and  let  a,  (3,  y  be  the  angles  OP 
makes  with  the  axes.     Then  the  equation  of  the  plane  is 

x  cos  a  +  y  cos  p  +  z  cos  y  —  p. 

The  equation  of  a  plane 
whose  intercepts  on  the  axes 
are  a,  b,  and  c  is : 


(9) 


a      b     c 


1. 


Fig.  80 


The   general   equation   of 
the  first  degree : 

(10)    Ax  +  By+Cz  +  D  =  0 

can  be  thrown  into  the  form 
(8)  as  follows : 

C       -D 


—  X  +  —  V  +  —Z 

A        A*      A  A    ' 

where  A=  (A2  +  B2+  C2)*.  If  D  was  originally  positive, 
change  the  signs  of  all  the  coefficients,  so  that  D  become 
negative:   —  DgrO.     Then 


(12) 


cosa  =  — ,      cos/8 
A'  h 


B 


C 

COS  y—  — . 
7       A" 


p 


D 


For  most  purposes  it  is  sufficient  to  note  that 


PARTIAL   DIFFERENTIATION  285 

(13)  cos  a :  cos  /? :  cos  y  =  A :  B :  C. 
The  angle  between  two  planes : 

Ax  +  By+Cz  +  D  =  0, 
A'x  +  B'y+C'z  +  D'  =  0, 
is  given  by  the  formula : 

(14)  q089  =  AA'  +  BB'  +  GC'. 

V     ;  AA' 

The  planes  are  perpendicular  if 

(15)  AA'  +  BB'+CC'  =  0, 
and  conversely.     They  are  parallel  if 

(16)  A:A'  =  B:B'  =  C:C'. 

The  distance  d  of  the  point  P:  (x1}  ylt  z^)  from  the  plane  (8)  is 

(17)  d  —  ±  {xx  cos  a  +  yx  cos  /?  +  zx  cos  y  —p), 

where  the  lower  sign  is  to  be  used  if  0  and  P  are  on  the  same 
side  of  the  plane,  and  the  upper  sign  in  case  they  are  on  oppo- 
site sides. 

The  Straight  Line.   A  straight  line  may  be  determined  (a) 
as  the  intersection  of  two  planes : 

/18n  f  Ax  +  By  +  Cz  +  D  =  0, 

K     }  \A'x+B'y  +  C'z  +  D'  =  0'y 

(b)  by  its  direction  and  one  of  its  points : 

(19)  «— «fe.-y-yo_*— %. 

cos  a       cos  /?      cos  y ' 

(19a)   °r  x-^^y-y.^zj-z,^ 

I  m  n 

where  l:m:n  =  cos  a :  cos (3 : cos y ; 

(c)  by  two  of  its  points : 


^-xo     yx-y0     zx-  z0 
In  the  latter  case 


(20) 


286  CALCULUS 

(21)  cos  a :  cos  /? :  cos y  =  Xj  —  x0 :  y1  —  y0 :  zx  —  z0. 

If  (x0,  y0,  z0)  is  a  point  of  the  line  (18),  the  equations  may 
be  expressed  in  the  form  (19)  as  follows : 

/29\  x  —  x0     =     y  —  y0     =      z  —  z0 

V";  \  B     G    \    "    \   C    A    \    "    \  A    B    \    ' 

I  B'    C   |  I   C   A'  |  I  A'   B'  I 

The  direction  cosines  of  (18)  are  thus  given  by  the  relations : 

(23)  cos«:cos0:eosy  =  |  f,   g,  |  :  |  g,  ^  |  :  |  ^  f,  |  • 

If  the  line  is  given  as  the  intersection  of  two  planes  perpen- 
dicular respectively  to  the  x,  y  and  the  x,  z  planes  :  * 

(24)  y  =px  +  6,  z  =  qx  +  c, 

its  equations  can  be  brought  into  the  form  (19)  as  follows : 
x  —  0  _y  —  b  _z  —  c 


(25) 
Hence 

(26) 


1  p  q 

1 


cos  a  = 


VI  -i-p2  +  q2 
cos  /?  = p  , 

Vi+^  +  g8 


COSy 


Vi+^  +  g2 


.Line  Normal  to  a  Plane.     The  equations  of  a  straight  line 
passing  through  any  point  (x0,  y0,  zQ)  of  space  and  perpendicular 
to  the  plane  (18)  are : 
/97\  x  —  xq  _  y-y0  _  z-z0 

Plane  Normal  to  a  Line.     The  equation  of  a  plane  passing 
through  any  point  (xQ,  y0,  z0)  of  space  and  perpendicular  to  the 
line  (19  a)  is : 
(28)  i(x  —  x0)  +  m(y  —  yQ)  +  n(z  —  z0)  =  0. 

*The  p  that  figures  here  has,  of  course,  nothing  to  do  with  the  former 
p,  the  length  of  the  perpendicular. 


PARTIAL   DIFFERENTIATION  287 

Variable  Plane  through  a  Line.     The  equation  of  a  variable 
plane  through  the  line  (18)  is : 

(29)     (Ax  +  By  +  Cz  +  D)+k(A'x  +  B'y  +  C'z  +  iy)  =  0, 

where  A;  may  have  any  value  whatever. 

Three  Planes  through  a  Line.     The  condition  that  the  three 
planes 

Ax  +  By  +  Cz  +  D  =  0, 

A'x  +  B'y+C'z  +  D'  =  0, 

A"x  +  B"y+C"z  +  D"  =  0, 

all  intersect  in  one  and  the  same  straight  line  is  that 


(30) 


B 

C 

/ 

B< 

C 

n 

B" 

C" 

o, 


and  that  they  have  one  point  in  common. 

The  student  should  notice  that,  while  one  equation  deter- 
mines a  plane,  it  always  takes  two  equations  in  x,  y,  z  to 
determine  a  line. 

'  3.  Partial  Derivatives.  If  in  the  function  (1)  we  hold  y  fast 
and  differentiate  with  respect  to  x,  we  obtain  the  partial  deriva- 
tive of  z  with  respect  to  x,  denoted  by 


£        or        f9{x,y). 

Similarly, 

differentiation  with  respect  to  y,  x  being  constant, 

gives 

|        or        f,(*,p), 

Thus  if 

z  =  e~x  sin  y} 

dz__ 

dx 

—  e~xs'my,            7^  =  e~xcosy. 
dy 

288  CALCULUS 


EXERCISES 


1. 

Find 

dx 

dz 
and     —  in  each  of  the 
dy 

following 

cases : 

(a) 

z  =  x\ogy; 

(P) 

z  =  ax2-\-2bxy-\-cy2 

5 

M 

e*y 

*?  +  y2' 

(d) 

tf  +  y*  +  z2  =  a2. 

2. 

Find  all  the  partial  derivatives  of  u, 

when  u  =  ax  +  by  +  cz. 

3. 

If 

pv  =  apQvQ  T, 

where  a,  p0>  ^o  are  constants,  find 

4.  Geometric  Interpretation.  Geometrically  the  meaning  of 
the  partial  derivatives  in  case  there  are  but  two  independent 
variables  is  as  follows.  Holding  y  fast  is  equivalent  to  cut- 
ting the  surface  (1)  by  the  plane  y  =  yQ.  The  section  is  a 
plane  curve :  */       '\ 

and  — -  is  the  slope  of  this  curve.     Similarly  —■  is  the  slope  of 
the  curve  ~t        N 

*  =/Oo,  y)- 

We  thus  have  the  slopes  of  two  tangent  lines  to  the  surface 
(1)  at  the  point  (x0,  y0,  z0),  and  hence  we  can  readily  determine 
the  equation  of  the  tangent  plane  through  this  point.  For  the 
tangent  plane  at  a  point  contains  all  the  tangent  lines  at  the 
point  and  is  determined  by  any  two  of  them.  If,  therefore, 
we  write  the  equation  of  the  tangent  plane  with  undetermined 
coefficients  in  the  form : 

z  -  z0  =  A  (x  -  x0)  +  B  (y  —  y0), 

we  have  only  to  require  that  the  slope  of  the  line  in  which 
this  plane  is  cut  by  the  plane  y  =  y0,  i.e. 


PARTIAL  DIFFERENTIATION 


289 


z  —  z0  =  A  (x  —  x0) 

be  dz/dx,  formed  for  the 
point  (sc0,  2/0),  —  we  will 
denote  this  quantity  by 
(dz/dx)Q,  —  and  similarly 
that  the  slope  of  the  line  in 
which  the  plane  is  cut  by 
the  plane  x  ==  x0 : 

z-z0  =  B(y-y0), 

be  (dz/dy)Q.     Hence 


dxja 
and  we  obtain  as  the  equation  of  the  tangent  plane : 

(31)  Z-2»  =  (l)0(a;-^  +  (|)0^-^ 

From  (28)  it  follows  that  the  equations  of  the  normal  line 
(or  simply  the  normal)  to  the  surface  (1)  at  the  point  P: 
0»o>  Vo,  3o)  are: 
(32) 


(-)        (-) 
\dxj0       \dyj0 


The  direction  cosines  of  the  normal  are  given  by  the  relations 
(33)  cos«:cos0:cos7  =  (!)o:(!);-l. 


EXERCISES 


Find  the  equations  of  the  tangent  plane  and  the  normal  to 
the  following  surfaces : 

1.  z  =  tan-1?. 

x 

Ans.   y0x  —  x0y  +  (x02  +  y02)(z  —  zQ)  =  0; 
Vo      ~  —Xq       «o2  +  2/o2' 


290  CALCULUS 

2.  z  =  ax2  -f  by2. 

Ans.    For  the  tangent  plane :  z  =  2ax0x  +  2 6^^  —  2o» 

3.  «2  4-  y2  +  a;2  =  a2. 

4.  Show  that  the  surface 

z  =  xy 

is  tangent  to  the  x,  y  plane  at  the  origin. 

,   5.   The  sphere:  ^y+>.ii 

and  the  ellipsoid:         3^  +  ^  +  ^  =  20 

intersect  in  the  point  (—1,  —  2,  3).     Find  the  angle  at  which 
they  cut  each  other  there.  Ans.   23°  33'. 

6.  What  angle  does  the  tangent  plane  of  the  ellipsoid  in  the 
preceding  question  make  with  the  x,  y  plane  ?        Ans.   59°  2'. 

7.  At  what  angle  is  the  surface 

z  as  Sxy2  —  5x2y  —  7x  +  3y 
cut  by  the  axis  of  x  at  the  origin  ?  Ans.   65°  41 '. 

'    5.    Derivatives  of  Higher  Order.     The  first  partial  deriva- 
tives of  the  function 

are   themselves   functions   of  sc  and  #,   and   can  in  turn   be 
differentiated : 

d2u      *   ,       n  /.  /       \  0  /dw\        d2it        ^  /       N      , 

It  can  be  shown  that  the  order  of  differentiation  does  not 
matter,  provided  merely  that  the  derivatives  concerned  are 

continuous  functions : 

d2u        d2u 


(34) 


cxdy      dydx 


PARTIAL  DIFFERENTIATION  291 

The  theorem  holds  for  functions  of  any  number  of  variables.* 
Let  us  verify  the  theorem  in  some  special  cases. 

(a)  u  —  excos  y ; 


du  r  •  d  fdu 

=  -e*smy, 
dy 


(P) 


u 


d  fdu\  „  • 

Hfe)— "-»■ 

a?  log  2  # 


y 

d2u 

1 

dzdx 

yz 

d2u 

_1^ 

du__\ogz 

dx        y 

du  _  x 
-  fo      2/z'  dscdz     yz 

EXERCISES 
/ 

1.  Verify  the  theorem  for  the  other  two  pairs  of  cross  deriv- 
atives in  (6). 

2.  Verify  the  theorem  in  each  of  the  following  cases: 
(a)     u  =  z  sin  xy ;         (b)     u  —  log  (xy2)  ;         (c)  w  =  ys. 

d3u         d*u  ffu 


0  3.    Prove  that 


dx2dy     dydx2     dxdydx 


f    4.   If  w  =  logV#2-f  y2, 

then  **  +  *•** 


dx2      dy2 

*  The  proofs  of  this  theorem  formerly  given  are  not  rigorous.  For  a 
critique  of  these  see  Gibson,  Elementary  Treatise  on  the  Calculus,  §  93, 
where  a  correct  proof,  due  to  Schwarz,  is  to  be  found  ;  or  Goursat- 
Hedrick,  Mathematical  Analysis,  vol.  1,  §  11.  A  simple  proof  can  be 
given  by  integration  ;  cf.  Whittemore,  Bulletin  Amer.  Math,  fioc,  ser.  2, 
vol.  4  (1898),  p.  389. 


292  CALCULUS 

1 
'   5.   If  u  = 


V  x2 -j- y2 -\-z2 

then  ^  +  ^  +  ^  =  0. 

dx2^  dy2  T  dz2 

1    6     If  —  =  *tH         and         —  =  —  — 

0sb.     %  dy  dx 

then  ^  +  ^  =  0. 

v  6.  The  Total  Differential.  Let  us  form  the  increment  of  the 
function  „,       N 

Au  =f(x0  +  A#,  y0  +  Ay)  -/(#0,  2/0)- 

If  we  subtract  and  add  the  quantity  f(x0,  y0  4-  Ay),  we  shall 
have: 

Aw  =f(x0  +  Ax,  y0  +  Ay)  -f(x0,  y0  +  Ay) 

+/Oo,  2/o  +  Ay)  -/(«<,,  2/0). 

Applying  the  law  of  the  mean  to  these  two  differences  gives : 

(35)         Au  =  Axfx  (x0  +  6 Ax,  y0  +  Ay)  +  Ayfy  (x0 ,y0  +  0'  Ay). 

Now  if  fx(x,  y)  andi/y(a;,  y)  are  continuous  functions  of  x,  y, 
fx(x0  +  6Ax,  y0  +  Ay)  will  approach  fx(x0,  y0)  as  its  limit  when 
Ax  and  Ay  both  approach  zero,  and  hence  will  differ  but  slightly 
fromXC^o?  2/o)  when  Ax  and  Ay  are  numerically  small: 

/x(z0  +  OAx,  y0  +  Ay)  =fx (z0,  y0)  +  e, 

where  c  is  infinitesimal : 

lim      e  =  0. 

Ax  =  0,  Ay=0 

Similarly,  the  limit  of/y(#0,  yo  +  0'Ay)  isfy(x0,  y0)  and 

/y  0»o,  2/o  +  0'Ay)  =/y Oo,  y0)  +  iy, 

where  rj  is  infinitesimal. 

Hence  (35)  may  be  written  in  the  form : 


PARTIAL  DIFFERENTIATION  293 

(36)  Aw  =  ^  Az  + |*  Ay  +  eAz  +  ,7  Ay, 

where  we  have  dropped  the  subscripts  and  replaced  fx(x,  y), 
fy(x,  y)  by  the  alternative  notation. 

Formula  (36)  is  analogous  to  the  second  formula  on  p.  92, 
and  so  it  is  natural  to  describe  the  linear  terms : 

du  .      ,  du  A 

—  &x  +  —Ay 
ex  cy 

as  the  principal  part  of  An.     The  remaining  terms  form  an 
infinitesimal  of  higher  order.* 

Definition.  We  define  the  total  differential  of  u  as  the 
principal  part  of  Aw : 

(37)  *-gA.  +  gA* 

Since  this  definition  holds  for  all  functions  u,  we  may  in 
particular  set  u  =  x.     From  (37)  follows  then  that 

(38)  dx  =  Ax. 
Similarly,  setting  u  =  y,  we  get : 

(39)  dy  =  Ay. 
Substituting  these  values  in  (37)  gives 

*  If  £  is  an  infinitesimal  depending  on  several,  let  us  say  two,  inde- 
pendent variables,  a  and  |8,  and  if  we  take  these  variables  as  the  princi- 
pal infinitesimals,  then  f  is  said  to  be  an  infinitesimal  of  higher  order  than 
a  and  /3  if 

lim  f         =  0. 

£  is  said  to  be  of  the  same  order  if 

K<—L—<G, 

where  J5T  and  #  are  constants,  both  positive  or  both  negative.     Instead  of 
the  above  ratio  we  might  equally  well  have  used 

r 


294  CALCULUS 

(A)  .  fcTg*  +  g* 

The  definition  (37)  and  the  theorems  (38)  and  (39)  can  be 
extended  to  functions  of  any  number  of  variables.  Thus  if 
u  =/(%,  y,  z)  we  have  by  definition 

,       du  A     .  du  A     .  du  A 
dw=  —  A#  +  7-  Ay  +  5-  A«, 
d#  dy  ^ 

and  we  conclude  as  above  that 

,   du  ,     .  du  ,     .  du   , 
aw  =  7—  ax  +  —  ay  +  tt  cfe. 

It  is  sometimes  convenient  to  use  the  partial  differentials  of 
u  obtained  by  allowing  only  one  of  the  variables  to  change : 


We  have  then: 

,          du  A        du  , 
dxu  =  — - -  Ax  =  —  ax, 

dx           ex 

etc. 

(40) 

du  =  dxu  +  dyu  4-  •  • 

•. 

Geometric  Interpretation.     In  the  case  of  a  function  of  two 
independent  variables : 

•—/fay)* 
the  increment  and  the  differential  of  the  function  admit  a 
simple  geometric  interpretation.  If  we  pass  a  plane  through 
the  point  P  :  (x0,  y0,  z0)  parallel  to  the  x,  y  plane  and  then 
draw  a  line  parallel  to  the  z-axis  and  cutting  that  plane  in  the 
points  x  =  x0  +  Ax,  y  =  y0  +  Ay,  the  segment  of  this  line  be- 
tween the  above  plane  z  —  z0  and  the  surface  is  (see  Fig.  81) : 

LQ  =f(x0  +  Ax,  y0  +  Ay)  -f(x0,  y0)  =  Az. 
The  equation  of  the  tangent  plane  at  P  is 


KIOo^KtV^' 


and  the  segment  of  the  line  between  the  plane  z  =  z0  and  this 
plane  is 


PARTIAL  DIFFERENTIATION  295 

The  difference : 

Az  —  dz  =  MQ  =  e  Ax  +  r}  Ay, 

is  an  infinitesimal  of  higher  order  than  Ax  and  Ay. 

■  7.  Continuation.  Change  of  Variable.  In  the  foregoing 
paragraph  we  have  assumed  that  x  and  y  are  the  independent 
variables.     If  each  depends  on  a  third  variable,  t : 

(41)  *  =  <Kt),         y  =+(t), 

then  u  becomes  a  function  of  a  single  variable,  t,  and  the  differ- 
ential of  such  a  function  has  already  been  defined,  Chap.  V,  §  4 : 

(42)  du  =  DtuAt  =  Dtudt. 
Also: 

(43)  dx  =  Dtx  dt,  dy  =  Dty  dt. 

Here  dt  =  At ;  but  da?  and  dy  are  not  in  general  equal  to  Ax 
and  Ay  respectively.  The  question  therefore  arises  :  Will  the 
theorem  (A)  still  hold  ?     Wo  proceed  to  show  that  it  will. 

Let  Ax  and  Ay  be  the  increments  that  x  and  y  receive  by 
virtue  of  (41)  when  t  has  the  increment  At.  Then,  substituting 
these  values  in  (36),  we  get  the  increment  of  u.  Now  divide 
through  by  At  and  take  the  limit  of  each  sidd: 

At=o\AtJ     cxAt=o\AtJ     dy*t±o\AtJ     ^t=o\  At         At  J 
The  last  limit  has  the  value  0,  and  hence 

(44)  Dtu  =  d^Dtx+d-^Dtyy 

ex  dy 

and  du  =  -^  dx  +  -^  dy.  q.  e.  d. 

ex  dy 

Thus  (A)  is  seen  to  hold  even  when  t  is  the  independent 
variable. 

Finally,  let  x  and  y  depend  on  r  and  8 : 

(45)  x  =  cf>(r,  s),  y  =  ^(r,  s). 


296  CALCULUS 

If  we  hold  s  fast  and  allow  r  alone  to  vary,  we  have  the  case 
just  treated,  the  independent  variable  now  being  r  instead  of  t. 
Hence  (44)  is  still  valid,  the  derivatives  with  respect  to  r  now 
being  partial : 

/to  du _dudx     du  dy 

dr      dx  dr     dy  dr 
In  like  manner : 

du _dudx     du  dy 
ds      dx  ds      dy  ds ' 

Let  us  state  this  result  in  the  form  of  a  theorem.  It  is  ap- 
plicable to  functions  of  any  number  of  variables. 

Theorem  1.     If 

and  if  each  of  the  arguments  x,  y  z,  •  •  •  is  made  to  depend,  on 
r,  8,  •  •  •  ; 

x  =  cf>(r,  s,  •  •  •),       y  =  «/'(r,  «,•••)>         z  =  <o(r,  s,  .  .  •), 

then,  if  all  the  derivatives  involved  are  continuous : 

x-px  du  _dudx     du  dy     dudz 

dr      dx  dr      dy  dr      dz  dr 

with  similar  formulas  for  — ,  etc.,  obtained  from  (B)  by  replacing 

r  by  s,  etc. 

The  number  of  variables  in  each  class,  (x,  yf  z,  •••)  and  (r,  s,  •••), 
is  arbitrary.  If,  in  particular,  there  is  only  one  variable,  x,  in 
the  first  class,  but  several  in  the  second,  we  have 

du  _  du  dx^ 
dr      dxdr' 

and  if  there  is  only  one  variable,  t,  in  the  second  class,  but  sev-  s 
eral  in  the  first,  then  we  have  formula  (44) : 

da  _  du  dx     du  dy  .  du  dz 
dt      dx  dt     dy  dt      dz  dt 


PARTIAL  DIFFERENTIATION  297 

Example.     Let 

u  =  exy, 

x  =  log  Vr2  +  s2,  y  =  tan"1-. 

r 

Th  pn  —  —  w«v  — -  —  xpw 

dx~V     '  dy  ' 

dx  __      r  dy  _    —  s 

and  hence  |=fef^ 

#r       rl  +  s2 

from  which  expression  x  and  ?/  can  be  eliminated  if  desired. 


EXERCISES 

si.  if 

u  =  x2  —  y2 

and 

J         x=  2r-3s  +  7, 
1         y  =  -r  +  8s-9, 

find  p. 
or 

Ans.     ^-  = 

Ux+2y. 

2.    In  the 

preceding  question,  find  -^. 

OS 

3    If 

u  =  iCl/* 

and 

x  = 

=  a  cos  0,      y  —  a  sin  0, 

z  =  b0, 

,find^' 
d$ 

V  4.    If 

x  +  y 
1  — sty 

and 

X- 

=  tan(2r— s2),          y  = 

cot  (r*s), 

find  ~  and 
or 

du 

5.   If 

u=f(*,y,z) 

and 


298 


CALCULUS 


x  =  ax'  4-  by'  -f  czf, 
2/=  «'#'+  6y+  c'z', 
z  =  a"rf  +  b"y'+c"z'i  J 


show  that 


and  find  — -  and 


du 

dx1 


du 
dx 


tldu 
dy 


,,du 


dy1 


'6.    If 

show  that 


dz' 

x  =  r  cos  <£, 


y  =  r  sin<£, 


WW    W    A^A 


Suggestion.     Compute  first  — -  and  —  in  terms  of  — -  and—. 

8.  Conclusion.  We  are  now  in  a  position  to  show  that 
the  theorem  (A)  is  true  no  matter  what  the  independent 
variables  are.     If 

and  '        x  —  4>  (r,  s),  y=t(r>  s)> 

then,  by  the  definition  (37), 


du 


du  .      ,  du  . 


Also 


Hence 


dx=iAr+8iAs> 
dy=iAr+diAs- 

du  7     ,  du  , 
0a>  ty 


(dud*  ,dudy\  *r,(?ufo  +  dJ:M\±s 
\fo dr      dy  drj  [dx^ds      B^ds) 


du  .      ,  du  .         , 


q.  e.  d. 


PARTIAL  DIFFERENTIATION  299 

We  will  state  the  result  as 

Theorem  2.     If 

u=f(?,y,z,  •  •  •), 

and  if  each  of  the  arguments  x,  y,  z,  •  •  •  is  made  to  depend  on 
r,  s,  •  •  •  : 

*= $(?>*> •  •  •)>      y—^(rt s>  •  -  •)>      * =  w (*j s>  •  •  •)> 

then,  if  all  the  first  partial  derivatives  are  continuous,  we  shall 
have : 

tfa;  (72/  ^ 

no  matter  whether  the  independent  variables  are  x,  y,  z,  •  •  •  or 
r,  s,  •  •  -. 

The  number  of  variables  in  each  class,  (x,  y,  z,  •  •  •)  and 
(r,  s,  •  '  •),  is  arbitrary. 

It  is  readily  shown  that  the  general  theorems  relating  to 
the  differentials  of  functions  of  a  single  variable : 

d(cu)  =  cdu, 

d  (u  +  v)  =  du  +  dv, 

d(uv)  =  udv  +  vdu, 

'u\      vdu—udv 


d 


V  V 


hold  for  functions  of  several  variables.  Moreover,  the  differ- 
ential of  a  constant,  considered  as  a  function  of  several 
variables,   is   0: 

dc  =  0. 

Example.     Let  us  work  the  example  of  §  7  by  means  of 
the  above  theorem. 

du  =  yexydx  +  xexvdy, 

r  & 

dx =   0  .    0 dr + 


r2  +  s2  ■    ^+^         > 
g  y 

*     r2  -f  s2         r  -f  s2 


300  CALCULUS 

Hence  du  =  ry~8fe>*dr  +  sy  +  ™e**ds 

r2  +  s2  r^  +  s2 

du  7    ,  du, 
=  —dr-\-—-ds. 

dr  ds 

Now  dr  =  Ar  and  ds  =  As  are  independent  variables,  and 
consequently  we  can  equate  their  coefficients  on  the  two  sides 
of  the  last  equation  :  * 

du  _ry  —  sx  xy  du _sy  +  rx 

dr~  ?  +  s2      '  ds"  r*  +  s2 

EXERCISES 

1.  Work  the  first  four  exercises  at  the  end  of  §  7  by  the 
method  just  explained. 

2.  If  u=f(x  +  a,y  +  b), 

,        .*  du     du  ,        du     du 

show  that  —-  =  —         and        —-  =  —-. 

dx      da7  dy      db 

3.  If  u=f(x)       and       x  =  3r  +  2s  +  7t, 

show  that  —  =  2  — . 

cs        dx 

9.  Euler's  Theorem  for  Homogeneous  Functions.  A  function 
u  is  said  to  be  homogeneous  if,  when  each  of  the  arguments 
is  multiplied  by  one  and  the  same  quantity,  the  function  is 
merely  multiplied  by  a  power  of  this  quantity.  For  definite- 
ness  we  will  assume  three  arguments : 

u=f(®,  y,  z), 
(46)  / O,  Xy,  Xz)  m  \»f  (x,  y,  z). 

*  The  reasoning  here,  given  at  greater  length,  is  as  follows.  Since  dr 
and  ds  are  both  arbitrary,  we  may  set  ds  =  0,  dr  =£■  0,  and  then  cancel  dr. 
Thus  the  coefficients  of  dr  on  both  sides  of  the  equation  are  seen  to  be 
equal.  Similarly,  setting  dr  =  0,  ds  ^t  0,  we  infer  the  equality  of,  the 
coefficients  of  ds. 


PARTIAL  DIFFERENTIATION  301 

The  exponent  n  of  A.  is  called  the  order  of  the  function. 
Thus  the  functions 

u  =  aa?+bxy  +  cy2,  u  =  J  log  (ar2  -f-  #2)  -  log  a, 

ax -{-by  z  ,      _,w 

ca  +  cty  -^2  _j_  y2  x 

are  homogeneous  of  order  2,  0,  0,  \,  1,  resoectively. 

If  in  particular  we  set  A.  =  -,  we  have 

x 

(47)  /<ftft«H#/fl.'.   -\ 

\      X      xj 

Let  the  student  verify  this  last  formula  for  each  of  the 
functions  above  given. 

Euler's  Theorem.  If  u  is  homogeneous  and  has  continuous 
first  partial  derivatives,  then 

/ns  du  ,     du  ,     du 

(C)  xyx+yYy+zTz=nu- 

We  have  by  (46) 

(48)  'f&ffi^-Vfto** 

xf  =  \x,     y'  =  \y,     z'  =  Xz. 
Differentiate  (48)  partially  with  respect  to  A. : 

(49)  fx(x\  y\  *>+/,(*',  y',  z')y+f(x',  y',  z')z  =  n\^f(x,yy  %% 

where  fx(x',  y',  zf)  denotes  as  usual  the  partial  derivative  of 
f(x,  y,  z)  with  respect  to  x,  the  arguments  being  subsequently 
replaced  by  x',  y',  z'  respectively.  If  we  now  put  A  =  l, 
(49)  assumes  the  form  (C),  and  the  theorem  is  proven. 

We  have  stated  and  proved  the  theorem  for  a  function  of 
three  variables.  But  theorem  and  proof  hold  for  a  function 
of  any  number  of  variables. 


302  CALCULUS 

EXERCISE 

Verify  Euler's  Theorem  for  each  of  the  above  examples. 

10.   Differentiation  of  Implicit  Functions.     Let  y  be  defined 
implicitly  as  a  function  of  x  by  the  equation  (cf.  Chap.  II,  §  9)  : 

(50)  F(x,y)=0. 
To  differentiate  y  we  begin  by  setting 

u  =  F(x,y) 
and  forming  the  total  differential  of  u : 

du  =  ?fdx  +  d-fdy. 
Cx  Cy 

This  relation  is  true,  no  matter  what  the  independent  variables 
are,  §  8,  Theorem  2.  We  may,  therefore,  in  particular  choose 
y  so  that  the  equation  (50)  is  satisfied.  Then  du  =  0,  and  we 
have: 

dF 

(51)  ^+M:^  =  o         or         4-_*L 
K     }  dx^dydx  dx  dF 

dy 
In  like  manner,  if  z  is  defined  by  the  equation : 

(52)  F(x,y,z)  =  0, 
we  can  differentiate  z  partially  by  setting 

u  =  F(x,  y,  z) 
and  taking  the  total  differential  of  each  side : 

,        dF  ,     .  dF,     ,  dF, 

du  =  — -  dx  -f  — -  dy  4-  — -  dz. 
cx  cy  cz 

This  equation  is  true,  no  matter  what  the  independent  variables 
are. 

If  in  particular  z  be  so  chosen  that  the  equation   (52)  is 
satisfied,  then  du  =  0,  and 


PARTIAL   DIFFERENTIATION  303 


<7a?  02/  (72 


But  dz  now  has  the  value : 


dz  =  —  d»  +  — -  cfa/. 

da;  dy 


Hence,  eliminating  dz,  we  have 

{W+W^fa,{£FdFdz\d   =Q 
\0a?       dzdx)  \dy       dzdy) 

Here  cto  =  A#  and   dy  =  Ay  are   independent  variables.     We 
may,  therefore,  set  dy  =  0,  dx  =£  0,  and  divide  through  by  cfcc : 


^qx  dF.dFdz      ft 

(53)  to+&fcT°' 


£? 

£2 


A  similar  equation  holds  for  dz/dy,  x  being  replaced  through- 
out in  (53)  by  y. 

The  student  should  notice  carefully  what  the  independent 
variables  are  in  each  differentiation.  Thus  dF  /dx  is  the 
derivative  of  a  function  of  three  independent  variables,  x,  y,  z, 
and  the  values  of  these  variables  are  not  in  general  such  as 
to  satisfy  the  equation  (52).  At  this  stage  of  the  work  (52)  is 
irrelevant,  does  not  exist  for  us,  has  not  as  yet  come  into  play. 
The  same  is  true  of  dF/dy  and  dF/dz.  When  we  come  to 
dz/dx,  however,  this  2  is  a  function  of  the  two  independent 
variables,  x  and  y,  —  and  such  a  function  that  (52)  is  satisfied. 

The  generalization  to  a  function  u  of  any  number  of  variables 
is  now  obvious : 

F(u,x,y,  z,  ■  •  .)  =  0, 

dF 

dFdu      dF_  du_       dx 

(54)  du  dz+~dx~°>  dx~~~dF' 

du 
etc. 


304  CALCULUS 

Example.     Differentiate  z  partially,  where 

a2^62     c2       ' 
Here 

and  we  have: 

a2^~&Yx~   * 


dx 

c2x  m 

dz 
dy 

c2y 
b2z 

b2       c2  dy        ' 

Several  Implicit  Functions.  We  may  have  two  implicit 
functions,  u  and  v,  of  any  number  of  variables,  x,  y}  •  •  •, 
defined  implicitly  by  two  equations  : 

1         <&(u,  v,  x,y,  -  -  -)  =  0. 

For  definiteness,  let  the  number  of  variables  x,  y,  •  •  •  be  two. 
Setting 

U  =  F(u,  v,  x,  y),  V=&(u,  v,  x,  y), 

and  taking  differentials,  we  have  :  i 


(56) 


,TT     dF  ,     ,  dF  ,     ,  dF,     .  dF  , 
d(7=  — -  c?m  -f  — -  dv  +  — -  dx  +  —  dy, 
cu  cv  ex  cy 

d  F  =  —  du  +  — -  d  v  +  -x-  daj  +  —  dy, 
cu  cv  ox  cv 


no  matter  what  the  independent  variables  are.  If  we  now 
require  that  u  and  v  be  so  determined  that  the  equations  (55) 
be  satisfied,  we  get:  d{7=0,  dV=0-?  and  furthermore  : 

du  =  -^  dx  +  -^  dy, 

dv  =  — -  dec  +  — -  dy. 
ex  cy 

Substituting  these  values  of  du  and  dv  in  the  right-hand  sides 


PARTIAL   DIFFERENTIATION 


305 


of  (56),  we  see  that  the  coefficients  of  dx  and  dy  are  equal  to 
0,  and  hence  we  get  the  two  equations  : 

dFBu     M^+^-aO 

du  dx      dv  dx      dx        ' 

f^^  +  ^^  +  ^-0 

du  dy      dv  dy      dy 


and  two  similar  equations,  in  which  F  is  replaced  by  S>. 
These  latter  equations  the  student  should  write  out  for  him- 
self. From  the  first  and  third  of  these  four  equations  we  can 
solve  for  du/dx  and  dv/dx,  and  from  the  second  and  fourth, 
for  du/dy  and  dv/dy.     Thus 


(57) 


du 

dx' 


F. 

K 

*x 

** 

F 

K 

*. 

with  similar  formulas  for 

dv/dx, 

du/dy,  dv/dy.     The  student 

should  also  write  these  out  clearly  and  neatly. 

The  generalization  is  now  obvious.     Thus  if 

F(u,  v,  w,x,y,  .  .  .)  =  0, 

(58)                   J         $  (u,  v,  w,  x,  y,  •  •  •)  =  0.. 

1         *(u,  v,w,  x,y,.-  .)=0, 

we  shall  have 

F     F     F 

<J>        cb        cb 

^*x        ^v        ^*w 

(59-)                        du  - 

\If         \b         \b 

*  X         T«         ^w 

-• 

W                         dx— 

F     F     F 

$«         ®v         ^«, 

*u       %        *«, 

The  determinant  that  appears  in  the  denominators  : 

dF    dF     dF 

dF    dF 

du     dv     dw 

du     dv 

d&     d$>     d<& 

(60) 

d&    ao 

du      dv 

y 

du     dv     dw 
d*    a*    d* 

i 

du 

dv     dw 

306  CALCULUS 

is  called  the  Jacobian  of  the  functions  F,  <$,  or  F,  3>,  #.  In 
the  foregoing  it  has  been  tacitly  assumed  that  all  the  partial 
derivatives  are  continuous  and  that  the  Jacobian  does  not 
vanish. 

11.   A  Question  of  Notation.     Problem.     Suppose 

u=f(x,y),  y  =  cf>(x,z), 

to  find  f*. 

ex 

Before  beginning  a  partial  differentiation  the  first  question 
which  we  must  ask  ourselves  is:  Wliat  are  the  independent 
variables  f  Hitherto  the  notation  has  always  been  such  as  to 
suggest  readily  what  the  independent  variables  are.  In  the 
present  case  they  may  be : 

(a)   x  and  y  ;   or   (b)   x  and  z ;  or    (c)   y  and  z. 
We  can  indicate  which  case  is  meant  by  writing  the  independ- 
ent variables  as  subscripts,  thus  : 

In  case  (c)  —  has  no  meaning. 

ex  : ' 

Another  notation  sometimes  employed  is  to  mark  the  vari- 
able or  variables  that  are  held  fast,  thus : 


dx  \:       w    dx 


Let  the  student  compute  -—  in  cases  (a)  and  (b). 

ox 

12.  Small  Errors.  In  the  case  of  functions  of  a  single  vari- 
able we  have  seen  that  the  linear  term  in  the  expansion  of 
Taylor's  Theorem : 

fix)  =f(x0)  +f'(x0)(x  -x0)  +  •  •  ., 

can  frequently  be  used  to  express  with  sufficient  accuracy  the 
effect  of  a  small  error  of  observation  on  the  final  result,  cf. 


PARTIAL  DIFFERENTIATION  307 

Infinite  Series,  §  27.  This  term,  /'  (x0)  (x  —  x0),  is  precisely  the 
differential  of  the  function,  df,  for  x  =  xQ. 

The  differential  of  a  function  of  several  variables  can  be 
used  for  a  similar  purpose.  If  x,  y,  •  •  •  are  the  observed  quan- 
tities and  u  the  magnitude  to  be  computed,  then  the  precise 
error  in  u  due  to  errors  of  observation  Ax  =  dx,  Ay  =  dy,  etc. 
is  Au.     But 

*,„!**+£*+. .. 

ex  cy 

will  frequently  differ  from  Au  by  a  quantity  so  small  that 
either  is  as  accurate  as  the  observations  will  warrant,  —  and 
du  is  more  easily  computed. 


Example.     The  period  of  a  simple  pendulum  is 

4 


T=2^!-. 


9 

To  find  the  error  caused  by  errors  in  measuring  I  and  g,  or  in 
the  variation  of  I  due  to  temperature  and  of  g  due  to  the  loca- 
tion on  the  earth's  surface.  # 

Here  dT=-±=dl-  £\fl  dg 

sllg         9^9 
or 

<W=l(tt_ldg 
T      2  1       2  g9 

and  hence  a  small  positive  error  of  &  per  cent  in  observing  I 
will  increase  the  computed  time  by  ^k  per  cent,  and  a  small 
positive  error  of  7c'  per  cent  in  the  value  of  g  will  decrease  the 
computed  time  by  £&'  per  cent. 

EXERCISES 

1.   A  side  c  of  a  triangle  is  determined  in  terms  of  the  other 
two  sides  and  the  included  angle  by  means  of  the  formula : 

c2  =  a2  +  b2  —  2ab  cos  w. 


308  CALCULUS 

Find  approximately  the  error  in  c  due  to  slight  errors  in  measur- 
ing a,  b,  and  o>. 

Ans.   The  percentage  error  is  given  by  the  formula : 

dc_(a  —  b  cos  <o)  da  +  (b  —  a  cos  <p)  db  +  ab  sin  w  dw 
c  a2-\-b2  —  2ab  cos  <d 

2.  Find  approximately  the  error  in  the  computed  area  of  the 
triangle  in  the  preceding  question. 

3.  The  acceleration  of  gravity  as  determined  by  an  Atwood's 
machine  is  given  by  the  formula  : 

2* 

Find  approximately  the  error  due  to  small  errors  in  observing 
s  and  t. 

4.  Describe  an  experiment  you  have  performed  to  determine 
the  focal  length  of  a  lens,  or  the  horizontal  component  of  the 
earth's  magnetic  force ;  recall  the  relative  degrees  of  accuracy 
you  attained  in  the  successive  observations,  and  discuss  the 
effects  of  the  errors  of  observation  on  the  final  result. 

13.   Directional  Derivatives.     Let  a  function 
w  =/(«,  y) 

be  given  at  each  point  of  a  region  8  of  the  x,  y  plane  and  let  a 
curve  C  be  given  passing  through  a  point  P:  (x0,  y0)  of  the 
^___^  region.     Let  P'  be  a  second  point  of  C 

f     s  \      and  form  the  quotient: 

/         J^       I  uP, -up 


PP' 


Fig.  82 

We  set  Up,  —  uP=  Au,  PP'  =  A£  and  write 

,  ■     An  _  du 


The  limit  of  this  quotient,  when  Pf 
approaches  P,  is  denned  as  the  direc- 
tional derivative  of  u  along  the  curve  C. 


PARTIAL   DIFFERENTIATION  309 

If,  in  particular,  C  is  a  ray  parallel  to  the  axis  of  x  and 

having  the  same  sense,  the  directional  derivative  has  the  value 

3u 
of  the  partial  derivative,  — ;  if  the  ray  has  the  opposite  sense, 

the  directional  derivative  is  equal  to  —  —-.    A  similar  remark 

dx 
applies  to  the  axis  of  y. 

To  compute  the  directional  derivative  in  the  general  case  we 

make  use  of  (36)  or  (37) ;  hence 

af =0  A£      cx\Ai=oA$J     dy\^=oA$J 

or 

//;ix  du     du  ,  du   . 

(61)  =       Cos  a  -f-  — -  sin  a. 

d$      dx  dy 

The  extension  of  the  definition  to  space  of  three  dimensions 
is  immediate.     We  have : 

//;o\  du     du  ,  du        r>  ,  du 

(62)  gi=^cos"+^co^+feco^' 

where  a,  /?,  y  are  the  angles  that  C  makes  at  P  with  the  axes. 


EXERCISES 

1.  If  a  normal  be  drawn  to  a  curve  at  any  point  P  and  if  r 
denote  the  distance  of  a  variable  point  of  the  plane  from  a 
fixed  point  0;  y,  the  angle  between  PO  and  the  direction 
of  the  normal,  show  that 

(63)  '  £  =  -<*>sy. 

2.  Explain  the  meaning  of  -^  and  show  that 

or 

(64)  dn^dr 
K     J  dr      dn 

14.   Exact  Differentials.     If  in  the  expression 

(65)  Pdx+Qdy 

P  and  Q  are  functions  of  x  and  y  subject  to  no  restriction  ex- 


310  CALCULUS 

cept  that,  along  with  whatever  derivatives  we  wish  to  use,  they 
be  continuous,  there  may  or  may  not  be  a  function  u  =f(x,  y) 
whose  total  differential : 

■,        du  ,     .  du  , 
du  =  —dx  +  —dy 

ex  cy 


coincides  with 

(65) 

.     If  there 

is  such  a 

function, 

then 

Now  since 

dx 

d_fdu\_ 
dy\dx)~ 

du  _ 
dy 

=  Q- 

we  see  that  P  and  Q  are  subject  to  the  restriction: 

dP=dQ 

v    '  dy      dx 

It  can  be  shown  that  conversely,  when  P  and  Q  do  satisfy 
(66),  there  always  does  exist  a  function  u,  of  which  (65)  is  the 
total  differential. #  In  this  case  the  expression  (65)  is  said  to 
be  an  exact  differential. 

Example.     Consider  the  expression : 

(2ax  +  by  +  l)dx  +  (bx  +  2cy  +  m)  dy. 

tt  dP     ,  dQ     , 

Here  —  =  b,  Tr  =  h> 

cy  ex 

and  hence  we  have  an  exact  differential  before  us.     To  inte- 
grate it,  begin  with 

^  =  P=2ax  +  by  +  l, 

ex 

and   integrate   each   side  with  respect  to  x,  regarding  y  as 
constant : 

u  =  ax2  +  bxy  +  lx  +  <f>  (y), 

the  constant  of  integration  depending,  of  course,  on  y.     Now 
differentiate  this  expression  for  u  with  respect  to  y : 

*Cf.  Goursat-Hedrick,  Mathematical  Analysis,  vol.  1,  §§  151,  152. 


PARTIAL  DIFFERENTIATION  311 

Comparing  this  last  expression  with 

Q  =  bx  +  2  ct/  -f-  m, 

we  see  that  <f>'(y)  =  2cy-\-m, 

<f>(y)  =  cy2  +  my+C. 

Hence  w  =  ax2  +  &#?/  +  cy2  +  Ix  +  m?/  +  O. 

If  we  have  three  independent  variables  and  the  expression 

Pdx+Qdy  +  Rdz, 

the  necessary  and  sufficient  condition  that  it  be  an  exact  differ- 
ential is  that 

(67)  dX=d3  dQ  =  dR  dR  =  dP 

dy      dx'         dz       dy '  dx       dz  ' 

It  is  assumed  that  the  partial  derivatives  are  continuous. 


EXERCISES 

Determine  which  of  the  following  expressions  are  exact  dif- 
ferentials and  integrate  such  as  are : 

1.  ( excos  y .  )  dx  —  (ex  sin  y  -f  7  sec2  y)  dy. 

\  VI  —  x2/ 

2.  (x  +  y)dx  +  (x-y)dy. 

3 .  yz  exyz  dx  +  zx  exy*  dy  •+-  xy  eyz  dz. 


312  CALCULUS 


EXERCISES 

1.  If  pv1Al=C,  find     ^. 

dp 

2.  If  u  =  ^l, 


X 


or 


■x  =  ri-s,         ,<1/  =  e'> 


3.    If  u  =  e*amv  +  xlog(x-\-y), 

x  =  pqr,  y  —  r  sin-1  (qr)  ; 


findf^. 
dq 


4.    If  u  =  2xy 

and  2#  +  32/-t-5z  =  l, 

explain  all  the  meanings  which  •—   may  have,  and  evaluate 

ex 

this  derivative  in  each  case. 


5.    If 

find*?. 
ox 


{u5  +  v5  4-  x5  =      3#, 
W3  +  v3_|_2/3=_3a.> 


6.    If  F=2mv 

dV 


and  j         ■*■+*  +  *-      %> 

US  +  VS  +  yS==_Sx> 

dx 

uev -\-vx  =  ysinu9 
u  cos  u  =  a?2  -+-  y% 


find- 

ax 


findf^ 
dy 


PARTIAL   DIFFERENTIATION  313 

8.    From  the  equations 

it  follows  that 

I  _  dx  du     dx  dv 

du  dx      dv  dx ' 

q__  dy  du  ,  dydv 

du  dx      dv  dx 


Explain  the  meaning  of  each  of  the  partial  derivatives.     Com- 

{x  SB  u  4-  vuv, 
y  =  v  —  uvu, 


du       j  du 

pute  _  and  — 

dx  dy 


/9.    If 

find  §*. 

dx 

'  10.   If  u  =  x2  -f  y2  +  z2        and        z  =  xyt, 

explain  all  the  meanings  of  —  • 

dx 

I        <l>(x,y)  =  0, 

dz  d<f>      dz  d(f> 

show  that  *  =  ^"frfe  . 

dx  d<f> 

dy 
12.    If  u=f(x  +  at,y  +  pt)y 

show  that  h.=  a*L+B^9 

dt  dx^Pdyf 

and  obtain  the  general  formula  for  -^  • 

dtn 


314 

CALCULUS 

13.    If 

u  =/(y  +  ax)  -f  <\>  (y  —  ax), 

show  that 

d2u _    2d2u 
dx2          dy2 

14.    If 

»-'(§: 

show  that 

du        du  _  r. 

dx.       dy 

15.    If 

u=f(x  +  u,  y-u), 

ex 

16.    If 

u=f(xu,y), 

find^. 

dx 

17.  Use  the  method  of  differentials  to  find  ~t  —  and  -^, 

dx    cy  c)t 

in  terms  of  /f  (£,  >?),/„(£  *?),  if 

w  =f(x  +  ut,  y  —  ut). 

18.  If  w  is  a  function  merely  of  the  differences  of  the  argu- 
ments xu  x2,  •  •  •,  xn  show  that 

du      du_  ,   t     t   ,    du  _  ~ 
dxx      dx2  dxn 

19.  If  u  and  v  are  two  functions  of  x  and  y  satisfying  the 

relations : 

du  __dv  du  _  _  #y 

dx      dy  dy  dx' 

show  that,  on  introducing  polar  coordinates : 

x  =  r  cos  <f>,  y  =  r  sin  <£, 
we  have 

du  __ldv  ldu  _       dv _ 

dr     r  d<j}'  r  d<f>  dr 


PARTIAL  DIFFERENTIATION  315 

20.  If 

f(x,  y)=0        and         <j>  (a?,  z)  =  0, 

show  that  *A%<ML  =  %*±. 

ex  dy  dz     ox  dz 

21.  If  4>(P>V,  0  =  °> 
show  that                         -£  —  _H  -s  _.  1. 

<7£  0V  (7/) 

Explain  the  meaning  of  each  of  the  partial  derivatives. 

22.  Under  the  hypotheses  of  question  19,  show  that 

d2u      ldu      1  d2u  _  ~ 

dt2,      r  dr      r2  dcf>2 

23.  If  u=f(x,  y)  is  homogeneous  of  order  n,  show  that 

<>d2u  .  o        d2u    ,     9d2u        t        -,x 
ar  — -  -f-  2  xy  7—-~  -f  V  tt-s  =  n(n  —  1 )  w. 
&»2  J  dxdy     *  dy2        y  J 

24.  If  w  is  a  function  of  x,  y,  z  and  x,  y,  z  are  connected  by 
a  single  relation,  is  it  true  that 

dy        dz    dy 

25.  If 

dU=6dS-pdv 

is  an  exact  differential,  and  if  S  and  v  can  be  expressed  as 
functions  of  the  independent  variables  6,  p,  show  that 

<W=_dp    •         M=_^ 

^      as'        ap      20' 

State  what  the  independent  variables  are  in  each  differentiation. 


CHAPTER   XV 

APPLICATIONS  TO  THE  GEOMETRY  OF  SPACE 

1.  Tangent  Plane  and  Normal  Line  to  a  Surface.  We  have 
already  obtained  the  equation  of  the  tangent  plane  to  the 
surface 

(1)  z=f(x,y) 

at  the  point  (xQ,  yQ,  z0)  in  Chap.  XIV,  §  4: 

(2)  .-^.g^-^  +  g^-^. 

Also  of  the  normal : 

/q\  x  —  x0  _  y-y0  _  z  —  z0 

{  }  (dj\  ~  f^\  ~~  -1  ' 

\dxj0      \dy)0 

If  the  equation  of  the  surface  is  given  in  the  implicit  form : 

(4)  F(x,y,z)=0, 

then  (2)  and  (3)  become  by  virtue  of  (53)  in  Chap.  XIV : 

/its  x-x0      y-y0      z-z, 

(6) 


(dF\  ~~  (W\       fd_F\ 
\dx)0      \dyj0      [dzjo 


For  the  direction  cosines  of  the  normal  at  (a?,  y,  z)  we  have, 
on  dropping  the  subscript  : 

(7)  cos  a  :  cos  B :  cos  y=  ■— -  :  ——  :  ■— • 

'      ex      dy       dz 

316 


APPLICATIONS   TO   THE   GEOMETRY   OF   SPACE     317 

Example.   Consider  the  ellipsoid  : 


Here 


t  +  t+tvel 

^(x-x0)  +  2^(y-y0)  +  2f(z-z0)  =  O 


a2  +  62       c2 
for  the  tangent  plane ;  and  for  the  normal : 

x0  Vo  *o 


EXERCISES 

1.  Find  the  equation  of  the  tangent  plane  and  the  normal 
of  the  cone : 

z2  =  2x2  +  1/, 
at  the  point  (2,  1,  3). 

Ans.   ±x  +  y-3z  =  0;     *^  =  y-l  =  ^zl. 

2.  How  far  distant  from  the  origin  is  the  tangent  plane  to 
the  ellipsoid : 

a?  +  3y2  +  2z2  =  9 

at  the  point  (2,  - 1,  1)  ?  Ans.    2.182. 

3.  Determine  the  angle  between  the  normal  to  the  ellipsoid 
in  the  preceding  question  at  the  point  (2,  —  1,  1)  and  the  line 
joining  the  origin  with  this  point. 

2.   Tangent  Line  and  Normal  Plane  of  a  Space  Curve.     A 

curve  in  space  may  be  given  analytically 

(a)  by  expressing  its  coordinates  as  functions  of  a  parameter: 

(8)  *»/(*),      y  =  <f>(t),      z=«K0; 

(6)  as  the  intersection  of  two  cylinders  : 

(9)  y  =  *(«),  *  =  <AO); 


318  CALCULUS 

(c)  as  the  intersection  of  two  arbitrary  surfaces : 

(10)  F(x,y,z)  =  0,  *(x,y,z)  =  0. 

A  familiar  example  of  (a)  in  the  case  of  plane  curves  is  the 

cycloid ;  also  the  circle.  In  the  case  of  space  curves  we  have 
the  helix : 

(11)  x  =  a  cos  0,  y  =  a  sin  0,         z  =  bO. 

This  curve  winds  round  the  cylinder  x2  +  y2  =  a2,  its  steepness 
always  keeping  the  same.  It  is  the  curve  of  the  thread  of  a 
screw  that  does  not  taper.  Again,  if  a  body  is  moving  under 
a  given  law  of  force  the  coordinates  of  its  centre  of  gravity 
are  functions  of  the  time,  and  we  may  think  of  these  as 
expressed  in  the  form  (a).  But  the  student  must  not  regard 
it  as  essential  that  we  find  a  simple  geometrical  or  mechanical 
interpretation  for  t  in  (a).     Thus  if  we  write  arbitrarily : 

(12)  x  —  log  t,         y  —  sin  t,         z  = 


<fl+t2 

we  get  a  definite  curve,  t  entering  purely  analytically. 

In  particular,  we  can  always  choose  as  the  parameter  t  in 
(a)  the  length  of  the  arc  of  the  curve,  measured  from  an 
arbitrary  point: 

(13)  x  =  f(s),       y  =  f(s),       •  =  *(«> 

The  form  (b)  may  be  regarded  as  a  special  case  under  (a), 
namely  that  in  which 

x  =  t. 

On  the  other  hand,  it  is  a  special  case  under  (c). 

The  Direction  Cosines.  To  find  the  direction  cosines  of  the 
tangent  to  a  space  curve  at  a  point  P:  (xQ,  y0,  z0),  pass  a  secant 
through  P  and  a  neighboring  point  P :  (x0  -f  Ax,  y0+Ay,  z0  +  Az). 
The  direction  cosines  of  the  secant  are : 

cosa'  =  -^:,  cos£'  =  -^-,  cosy'  =  =, 

PP  PP>  PP 

and  hence,  for  the  tangent, 


APPLICATIONS  TO   THE   GEOMETRY  OF   SPACE     319 


v        Ax        ,.       (Ax 
cos  a  —  lim  =  hm 


with  similar  formulas  for  cos  /?,  cos  y.     Hence 


PP'J 


(14) 


cos  a 


dx 


cos  /»  =  &, 


COSy 


dz 


ds  ds'  ,  '      ds 

Here  the  tangent  is  thought  of  as  drawn  in  the  direction  in 
which  s  is  increasing.  If  it  is  drawn  in  the  opposite  direction, 
the  minus  sign  must  precede 
each  derivative. 

From  (14)  it  follows  at  once 
that 
(15)     ds2  =  dx2  +  dy2+dz2. 

This  important  formula  can  be 
proven  directly  from  the  rela- 
tion 

ppi 2  =  Ax2  -f  Ay2  +  Az2. 

If  we  assume  the  form  (a), 

ds2  =  [f(t)2-hcf>'(ty  +  ^(t)2^dt2 


Fig.  83 


and 


(16) 


(17) 


cos  a  = 


f(t) 


VfW+fffl+rW 

= Jy/f  (}f + <t>'  (ff+V  (ty  at. 


cos/}  = 


COSv  = 


(18) 


Applying  these  results  to  (9),  we  get 

1 


cos  a 


Af    ^dx'^dx2 


etc., 


320 


CALCULUS 


(19) 


-/^♦s+s* 


The  Equations  of  the  Tangent  Line  and  the  Normal  Plane. 
For  the  tangent  line  we  have,  in  case  (a)  : 


(20) 

and  in  (6) : 

(21)     y-yo  = 


«z-_#o  =  y  —  yo  = « -  «o . 

f(t0)        *'(f0)        ffr)' 


(x-x0)9 


-z^(fX{x~xo)- 


\dxj0 

The  normal  plane  is  given  by 
(22)      f  (?.)  (x  -  x0)  +  *'  (t0) (y  -  y0)  +  ^  ft)  (i  -z0)  =  0 
in  (a)  ;  and  in  (6)  by 
fdyy 


(23) 


a-#o  + 


c?rc 


,^-*>-+(S>-*>-a 


On  the  other  hand,  the  tangent  line  in  case  (c)  may  be 
obtained  most  simply  as  the  intersection  of  the  tangent  planes 
to  the  surfaces  at  the  point  in  question  : 


(24) 


1    ©.<-*  >♦(&<»-*>+ 


These  equations  may  be  thrown  into  the  equivalent  form 

x  —  ^o     _     y  —  .Vo      _      g  —  zq      . 


Fy    Fz 

^       ^x 

^  *; 

%     *. 

0 

<*>*      <E>x 

0 

*x        % 

(25) 


Hence  we  see  that  the  direction  cosines  of  the  tangent  line 
to  the  curve  of  intersection  of  the  surfaces  (10)  are  given  at 
(x,  y,  z)  by  the  proportion : 

(26)      cos  «  .  cos  /?  :  cos  y  = 

The  equations  of  the  normal  plane  can  now  be  written  down  at 
once. 


1  F     F 

Fz    Fx 

K    Fs 

1       y         z 

^fz       ^x 

**    *, 

APPLICATIONS  TO   THE   GEOMETRY  OF   SPACE      321 


EXERCISES 

Find  the  equations  of  the  tangent  line  and  the  normal  plane 
to  the  following  space  curves  : 

1.  The  helix  (11)  and  the  curve  (12). 

2.  The  curve :        y2  =  2  mx,        z2  =  m  —  x. 

3.  The  curve:        2x2  +  3y2  +  z2  =  9,        ^  =  3^  +  ^, 
at  the  point  (1,  —  1,  2). 

4.  Find  the  angle  that  the  tangent  line  in  the  preceding 
question  makes  with  the  axis  of  x. 

5.  Compute  the  length  of  the  arc  of  the  helix : 

a?  =  cos0,        y  =  sir\0,        5z  =  0, 
when  it  has  made  one  complete  turn  around  the  cylinder. 

6.  How  steep  is  the  helix  in  the  preceding  question  ? 

7.  Show    that    the  condition   that   the   surfaces    (10)   cut 
orthogonally  is  that  i 


(27) 


dFd<j?  ,dFd&  ,  ^5$  =  0 
dx  dx      dy  dy      dz  dz 


8.    Show  that  the  condition  that  the  three  surfaces  : 

F(x,y,z)=0,         *(a>,y,2)=0,         *(x,y,z)=0, 

intersecting  at  the  point  (x0,  y0,  z0),  be  tangent  to  one  and  the 
same  line  there  is  that,  in  this  point, 

dF  dF  dF 

dx  dy  dz 

d®  d®  d® 

dx  dy  dz 

d*  d*  ?W 

dx  dy  dz 

It  is  assumed  that  in  no  row  do  all  the  elements  vanish. 


(28) 


=  0. 


322  CALCULUS 

9.  The  surfaces 

a?  +  y2  +  z2  =  3,        xyz  =  l,        z  —  xy, 

all  go  through  the  point  (1,  1,  1).     Find  the  angles  at  which 
they  intersect  there. 

10.  Obtain  the  condition  that  the  surface  (4)  and  the  curve 
(8)  meet  at  right  angles. 

11.  Find  the  direction  of  the  curve 

x  =  t2,        y  —  f,        z  =  t* 
in  the  point  (1,  1,  1). 

12.  Find  the  direction  of  the  curve 

xyz  =  1,  y2  =  x 

in  the  point  (1,  1,  1). 

13.  Find  all  the  points  in  which  the  curve 

x  =  t2,        y  =  P,        z  =  t* 
meets  the  surface 

z2  =  x  +  2y-2, 

and  show  that,  when  it  meets  the  surface,  it  is  tangent  to  it. 

14.  Show  that  the  surfaces 

cr     b*     & 
in  general  never  cut  orthogonally  ;  but  that,  if 

i+i-i-o 

a2  +  62     c*     - > 
they  cut  orthogonally  along  their  whole  line  of  intersection. 

15.  When  will  the  spheres 

^  +  ^  +  ^  =  1,  (x-a)2  +  {y-bf+(z-cy  =  l 

cut  orthogonally  ? 

16.  Two  space  curves  have  their  equations  written  in  the 
form  (13).  They  intersect  at  a  point  P.  Show  that  the  angle 
e  between  them  at  P  is  given  by  the  equation : 


APPLICATIONS  TO   THE   GEOMETRY  OF   SPACE     323 

cos  c  =  x[x[  +  y[y'2  +  afjaj, 

where  as{=^p,  etc. 

17.  The  ellipsoid:  cc2 -|- 3 2/2  -+-  2z2  =  9  and  the  sphere: 
252  _|_  y2  _|_  ^  _  g  mtersect  in  the  point  (2,  1,  1).  Find  the  angle 
between  their  tangent  planes  at  this  point. 

3.  The  Osculating  Plane.  Let  P :  (x0,y0,  z0)  be  an  arbitrary- 
point  of  a  space  curve  (8),  and  pass  a  plane 

(29)  A(x-x0)  +  B(y-  y0)  +  C(z-z0)  =  0 

through  P.     Then  the  distance  D  of  a  point 

P'l  x  =  f(t0  +  h),         y  =  cf>(t0  +  h),         z  =  if;(t0  +  h) 

of  the  curve  from  this  plane  will  be  in  general  an  infinitesimal 
of  the  first  order  with  reference  to  PP'  as  principal  infini- 
tesimal.    For 

±D  =  A(x-x0)  +  B(y-y())  +  C(z-z0) 

^A2  +  B2+C2 

where  x,  y,  z  are  the  coordinates  of  P'. 
Hence 

±D  =  A  ^ &  +  *>  ~/ft)1  +  *  C*ft>  +  *)  ~  *  W1  +  etc' . 
V^  +  Jff+C* 

Applying  Taylor's  Theorem  with  the  Eemainder  to  each 
bracket : 

f(to  +  h)-f(t0)  =  hf>(t0)+1£f<'(t(i  +  eh), 
etc., 


and  setting  V-424-  52  +  O2  =  A,  we  obtain 

±  D  =  A  r^/'ft)  +  B  +'  (O  +  0 «A '  (<b)]/A 

Hence 

lim  ±J?  =  Af'W  +  ^'to  +  CyW, 


324  CALCULUS 

and  this  will  not  =  0  if  A,  B,  C  are  chosen  at  random,  unless 
P  happens  to  be  a  point  at  which  /'  (t0),  <f>'  (t0)  if/'  (/0)  all  vanish. 
"We  exclude  this  case.  On  the  other  hand,  PP'  =  As  and  h  =  M 
are  infinitesimals  of  the  same  order,  since 


As 


lim  ^ = Dts = V/'  (t0y+  <t>'  (t0y + ^  (t0y  *  o. 

Thus  the  above  statement  is  proven. 

If,  however,  A,  B,  and  C  are  so  chosen  that 

(30)  Af'(t0)  +  B<f>'(t0)  +  Of  ft)  -  0, 

then  lim  ±  D/h  =  0  and 

,.     ±D     Af"(t0)  +  B<f>"(t0)  +  C<f,"(t0) 
2E*   h*   "  2A 

Now  (30)  is  precisely  the  condition  that  the  tangent  line  to  (8) 
be  perpendicular  to  the  normal  to  the  plane  (29),  and  hence 
the  tangent  will  lie  in  this  plane ;  i.e.  the  plane  (29)  is  here 
tangent  to  the  curve,  and  D  becomes  now  in  general  an 
infinitesimal  of  the  second  order.  But  if  A,  B,  and  C  are 
furthermore  subject  to  the  restriction  that 

(31)  Af"  (to)  +B<t>"(t0)  +  Cy'fo)  =  0, 

then  even  lim  ±  D/h2  =  0  and  D  becomes  an  infinitesimal  of 
still  higher  order ;  —  of  the  third  order,  as  is  readily  shown,  if 

Equations  (30)  and  (31)  serve  in  general  to  define  the 
ratios  of  the  coefficients  A,  B,  C  uniquely.  The  latter  may, 
therefore,  be  eliminated  from  (29),  (30),  and  (31),  and  thus 
we  obtain  the  equation  of  the  osculating  plane : 

x-x0    y-y0     z—z0 

(32)  f'(t0)     4>'(to)      «A'(*o)      =0. 

f"(t0)    4>»(t0)    f(to) 

The  osculating  plane  as  thus  defined  is  a  tangent  plane 
having  contact  of  higher  order  than  one  of  the  tangent  planes 


APPLICATIONS   TO   THE   GEOMETRY  OF  SPACE     325 

taken  at  random.  There  is  in  general  only  one  osculating 
plane  at  a  given  point.  But  in  the  case  of  a  straight  line  all 
tangent  planes  osculate.  Again,  if  f"(t0)  =  <f>"(t0)  =  <A"ft>)  =  ^> 
the  same  is  true.  The  osculating  plane  cuts  the  curve  in 
general  at  the  point  of  tangency;  for  the  numerator  of  the 
expression  for  ±  D  changes  sign  when  h  passes  through 
the  value  0. 

It  is  easy  to  make  a  simple  model  that  will  show  the  oscu- 
lating plane  approximately.  Wind  a  piece  of  soft  iron  wire 
round  a  broom  handle,  thus  making  a  helix,  and  then  cut  out 
an  inch  of  the  wire  and  lay  it  down  on  a  table.  The  piece  will 
look  almost  like  a  plane  curve  in  the  plane  of  the  table,  and 
the  latter  will  be  approximately  the  osculating  plane. 

The  normal  line  to  a  space  curve,  drawn  in  the  osculating 
plane,  is  called  the  principal  normal.  The  centre  of  curvature 
lies  on  this  line,  the  radius  of  curvature  being  obtained  by  pro- 
jecting the  curve  orthogonally  on  the  osculating  plane  and 
taking  the  radius  of  curvature  of  this  projection. 

If  a  body  move  under  the  action  of  any  forces,  the  vector 
acceleration  of  its  centre  of  gravity  always  lies  in  the  osculat- 
ing plane  of  the  path. 

When  the  equation  of  the  curve  is  given  in  the  form  (9),  the 
equation  (32)  becomes: 


EXERCISES 

1.  Find  the  equation  of  the  osculating  plane  of  the  curve 
(12)  at  the  point  t  =  w. 

2.  Find  the  equation  of  the  osculating  plane  of  the  curve  of 
intersection  of  the  cylinders  : 

sc2  +  y2  =  a2,  x*  4-  z2  =  a2, 

and  interpret  the  result. 


326  CALCULUS 

Suggestion.     Express  x,  y,  z  in  terms  of  t : 

x  =  a  cos  t,  y  =  a  sin  £,  2  =  a  sin  £. 

3.  Show  that  the  centre  of  curvature  of  a  helix  lies  on  the 
radius  of  the  cylinder  produced. 

4.  Show  that  the  osculating  plane  of  the  curve 

V  =  A  z2  =  l-y 

at  the  point  (0,  0,  1)  has  contact  of  higher  order  than  the 
second. 

4.   Conf ocal  Quadrics.  *     Consider  the  family  of  surfaces : 

where  X  is  a  parameter  taking  on  different  values.  Each  sur- 
face of  the  family  is  symmetric  with  regard  to  each  of  the  co- 
ordinate planes.  We  may,  therefore,  confine  ourselves  to  the 
first  octant. 

If  A.  >  —  c2,  we  have  an  ellipsoid,  which  for  large  positive 
values  of  X  resembles  a  huge  sphere.  As  X  decreases,  the  sur- 
face contracts,  and  as  X  approaches  —  c2,  the  ellipsoid,  whose 
equation  can  be  thrown  into  the  form : 


v    T   \       a2  +  X     b2  +  X/ 


*  No  further  knowledge  of  quadric  surfaces  is  here  involved  than  their 
mere  classification  when  their  equation  is  written  in  the  normal  form 

a2  X  b2  X  C2 

See  Bailey  and  Woods,  Analytic  Geometry,  p.  316.     It  is  desirable  that 
the  student  have  access  to  models  of  the  three  types  here  involved. 

The  student  should  work  out  for  himself,  after  a  first  reading  of  this 
paragraph,  the  corresponding  treatment  of  the  confocal  conies  in  the 
plane  : 


r'2 


+  =^-=1 


a2  +  X     62  +  X 


APPLICATIONS   TO  THE  GEOMETRY  OF   SPACE      327 

flattens  down  toward  the  plane  z  =  0  as  its  limit,  —  more  pre- 
cisely, toward  the  surface  of  the  ellipse 

In  so  doing,  it  sweeps  out  the  whole  first  octant  just  once,  as 
we  shall  presently  show  analytically. 

Let  A  continue  to  decrease.     We  then  get  the  family : 

(35)  ~iT-  +  ^T 7JV-\  =  1>    -&20<-c2. 

These  are  hyperboloids  of  one  nappe,  and  they  rise  from  coin- 
cidence with  the  plane  z  =  0  for  values  of  /a  just  under  —  c2, 
sweep  out  the  whole  octant,  and  flatten  out  again  toward  the 
plane  y  =  0  as  their  limit  when  /a  approaches  —  b2. 

Finally,  let  A  trace  out  the  interval  from  —  62  to  —  a2.  We 
then  get  the  hyperboloids  of  two  nappes : 

(36)  S- 7^ r A r-i  -a2<v<-b2. 

V     )  a2  +  v      -02  +  v)      -(c'  +  v) 

These  start  from  coincidence  with  the  plane  y  —  0  when  v 
is  near  —  62,  sweep  out  the  octant,  and  approach  the  plane 
x  —  0  as  v  approaches  —  a2. 

Theorem  1.  Through  each  point  of  the  first  octant  passes  one 
surface  of  each  family,  and  only  one. 

Let  P :  (x,  y,  z),  be  an  arbitrary  point  of  this  octant.  Then 
x  >  0,  y  >  0,  z  >  0.  Hold  x,  y,  z  fast  and  consider  the  function 
of  A: 

The  function  is  continuous  except  when  A  =  —  c2,  —  b2,  or  —a*. 
In  the  interval  —  c2  <  A  <  -f-  oo  we  have  * 

/(+»)= -1,         lim    /(A)  =+oo. 

A=-c*+ 

*  The  notation  lim  /(x),    lim  f(x)  is  explained  in  Chap.  XI,  §  9. 

x=a+  x=a— 


328  CALCULUS 

Hence  the  curve 

crosses  the  axis  of  abscissas  at  least  once  in  this  interval. 
On  the  other  hand 

ft  m ^ V z        ^  q 

J  K  J  (a2  +  A)2      (&2  +  A)2      C^  +  A)2       * 

Hence  /(A.)  always  increases  as  X  decreases,  and  so  the  curve 
cuts  the  axis  only  once  in  this  interval.  We  see,  therefore, 
that  one  and  only  one  ellipsoid  passes  through  the  point  P. 

Similar  reasoning  applied  to  the  intervals  (—  b2,  —  c2)  and 
(—a2,  —  b2)  shows  that  one  and  only  one  hyperbola  of  one 
nappe,  and  one  and  only  one  hyperbola  of  two  nappes  pass 
through  P. 

Theorem  2.  The  three  quadrics  through  P  intersect  at  right 
angles  there. 

The  condition  that  two  surfaces  intersect  at  right  angles  is 
given  by  (27).  Applying  this  theorem  to  (34), and  (35)  we 
wish  to  show  that 

2x       2x  2y       2y  2z       2z    =Q 

a?  +  ka2+  (j.     b2  +  \b2  +  f*.     c2  +kc2  +  n 

Now  subtract  (35)  from  (34)  : 

(     A       *        .        y2        i         ^2       Lq 

^       ;L(a2  +  A)(a2  +  /x)"h(62+A)(62  +  /.)"1"(c2  +  A)(c2  +  /A)J       ' 

and  since  /*  —  A  =£  0,  this  proves  the  theorem. 

The  three  systems  of  surfaces  that  we  have  here  investigated 
are  analogous  to  the  three  families  of  planes  in  cartesian  coor- 
dinates, to  the  spheres,  planes,  and  cones  in  spherical  polar 
coordinates,  and  to  the  planes,  cylinders,  and  planes  in  cylindri- 
cal polar  coordinates.  They  form  what  is  called  an  orthogonal 
system  of  surfaces,  and  enable  us  to  assign  to  the  points  of  the 
first  octant  the  coordinates  (A,  /*,  v),  where 

-c2<A<+oo,         _&2</jt<_c2j         _a2<v<-62. 


APPLICATIONS   TO   THE   GEOMETRY  OF   SPACE     329 

5.  Curves  on  the  Sphere,  Cylinder,  and  Cone.  In  order  to 
study  the  properties  of  curves  drawn  on  the  surface  of  a  sphere, 
we  introduce  as  coordinates  of  the  points  of  the  surface  the 
longitude  0  and  the  latitude  <£.  Any  curve  can  then  be  repre- 
sented by  the  equation 
(37)  F(0,  <l>)  =  0. 

To  determine  the  angle  w  between  this  curve  and  a  parallel 
of  latitude,  draw  the  meridians  and  the  parallels  of  latitude 
through  an  arbitrary  point  P :  (0O>  <£o)  and  a  neighboring  point 
P*  >  (#o  +  A0,  <£0  +  A<£)  of  this  curve.  We  thus  obtain  a  small 
curvilinear  rectangle,  of  which  the  arc  PP'  is  the  diagonal. 
We  wish  to  determine  the  angle 

<o  =  Z.MPP'. 

Now  consider,  alongside  of  the 
curvilinear  right  triangle  MPP'  a 
rectilinear  right  triangle  whose 
hypothenuse  is  the  chord  PP'  and 
one  of  whose  legs  is  the  perpen- 
dicular PM1  let  fall  from  P  on 
the  meridian  plane  through  P'. 
The  angle 

<1>'  =  ZM1PP' 

of  this  triangle  evidently  approaches  w  as  its  limit  when  P 
approaches  P. 


Fig.  84 


We  have 


tan 


MXP 
PMi 


Now  PMl  differs  from  PM=  a  cos  <£0  A0  by  an  infinitesimal  of 
higher  order  and  likewise  MXP  differs  from  MP'  =  aA<f>  by 
an  infinitesimal  of  higher  order.  Hence,  by  the  theorem  of 
Chap.  V,  §  2,  we  obtain : 

lim  tan  w'  =  lim  —J —  =  hm £ — -, 

P'±f  p'±p  PMX     A0=o  a  cos  <t>0  A0 


tan 


cos<£( 


De<t>, 


330  CALCULUS 

or,  dropping  the  subscript : 

(38)  tano>  =  -J—^. 
\ " '  cos  <\>  dO 

In  order  to  obtain  the  differential  of  the  arc  of  the  curve 

(37)  we  write  down  the  Pythagorean  Theorem  for  the  triangle 
PM^: 

PP'2  =  PMl2-{-M1P'\ 

divide  through  by  A02  and  then  let  A0  approach  0  as  its 
limit.  Since  the  chord  PP'  differs  from  the  arc  As  by  an 
infinitesimal  of  higher  order,  we  have : 

lim  f^]2  =  lim  f — Y=  a2cos24>0  +  a2 lim  f^Y 
p>±p\&0J       p'±p\A0j  ^  p>±p\A0J  , 

(Ddsy  =  a2cos2<f>  +  a2(De<t>)2, 

(39)  ds*  =  a2  [cos2<£d02  +  d<j>2]. 

Rhumb  Lines.  A  rhumb  line  or  loxodrome  is  the  path  of  a 
ship  that  sails  without  altering  her  course,  i.e.  a  curve  that 
cuts  the  meridians  always  at  one  and  the  same  angle.  If  we 
denote  the  complement  of  this  angle  by  o>,  then  we  have  from 

(38)  for  the  determination  of  the  curve : 

— £-  =  c!0tanG>, 
cos<f> 

(40)  <9tano>=   fJ±-  =  \ogten(±  +  Z)  +  a 
v     '  J  cos<£  \2      4y 

This  is  the  equation  of  an  equiangular  spiral  on  the  sphere, 
which  winds  round  each  of  the  poles  an  infinite  number  of 
times. 

EXERCISES 

1.  Show  that  the  total  length  of  a  rhumb  line  on  the  sphere 
is  finite. 


APPLICATIONS  TO   THE   GEOMETRY  OF   SPACE     331 

2.  The  cartesian  coordinates  of  a  point  on  the  surface  of  a 
sphere  are  given  by  the  equations  : 

x  =  a  cos  <£  cos  0,        y  =  a  cos  <f>  sin  0,        z  =  a  sin  <£ . 

Deduce  (39)  from  these  relations  and  the  equation : 

ds2  =  dx>  +  dy2  +  dz2. 

3.  Taking  as  the  coordinates  of  a  point  on  the  surface  of  a 
cone  (p,0),  where  p  is  the  distance  from  the  vertex  and  0  is 
the  longitude,  show  that 

(41)  tan-- 3&— . 


pdOsina 


4.  Obtain  the  equation  and  the  length  of  a  rhumb  line  on 
the  cone. 

5.  The  preceding  two  questions  for  a  cylinder. 

6.  Mercator's  Chart.  In  mapping  the  earth  on  a  sheet  of 
paper  it  is  not  possible  to  preserve  the  shapes  of  the  countries 
and  the  islands,  the  lakes  and  the  peninsulas  represented. 
Some  distortion  is  inevitable,  and  the  problem  of  cartography 
is  to  render  its  disturbing  effect  as  slight  as  possible.  This 
demand  will  be  met  satisfactorily  if  we  can  make  the  angle 
at  which  two  curves  intersect  on  the  earth's  surface  go  over 
into  the  same  angle  on  the  map.  For  then  a  small  triangle 
on  the  surface  of  the  earth,  made  by  arcs  of  great  circles,  will 
appear  in  the  map  as  a  small  curvilinear  triangle  having  the 
same  angles  and  almost  straight  sides,  and  so  it  will  look  very 
similar  to  the  original  triangle.  What  is  true  of  triangles 
is  true  of  other  small  figures,  and  thus  we  should  get  a  map 
hi  which  Cuba  will  look  like  Cuba  and  Iceland  like  Iceland, 
though  the  scale  for  Cuba  and  the  scale  for  Iceland  may  be 
quite  different. 

A  map  meeting  the  above  requirement  may  be  made  as  fol- 
lows. Regarding  the  earth  as  a  perfect  sphere,  construct  a 
cylinder  tangent  to  the  earth  along  the  equator.      Then  the 


332 


CALCULUS 


meridians  shall  go  over  into  the  elements  of  the  cylinder  and 
the  parallels  of  latitude  into  its  circular  cross-sections  as  fol- 
lows :  Let  P  be  an  arbitrary  point  on  the  earth,  Q>  its  image 
on  the  cylinder. 

(a)  Q  shall  have  the  same  longitude,  0,  as  P. 

(b)  To  the  latitude  <f>  of  P  shall  correspond  a  distance  z  of  Q 
from  the  equator  such  that  the  angle  w  which  an  arbitrary 
curve  C  through  P  makes  with  the  parallel  of  latitude  through 
Pand  the  angle  <o1  which  the  image  C1  of  C  makes  with  the 
circular  section  of  the  cylinder  through  Q  shall  be  the  same. 
Now  from  (38) 

tanco  =  —^-. 
dO  cos  <j> 

On  the  other  hand, 

dz 


tan 


(Dl  = 


adO 


Fia.  85 


Hence,  setting  a  for  convenience 
=  1,  we  get 

d*      =^     or     dz  =  M- 

dO  cos  cf>      dO 


cos  $' 


J  cos  <f> 


4> 


+ 


1). 


t     l0S""\2 

the  constant  of  integration  vanishing  because  z  =  0  corresponds 
to<£  =  0. 

Thus  a  point  in  latitude  60°  N.  goes  over  into  a  point  distant 
1.32  units  from  the  equator. 

The  cylinder  can  now  be  cut  along  an  element,  rolled  out  on 
a  plane,  and  the  map  thus  obtained  reduced  to  the  desired  scale. 

This  map  is  known  as  Mercator's  Chart.*  It  has  the 
property  that  the  meridians  and  the  parallels  of  latitude  go 
over  into  two  orthogonal  families  of  parallel  straight  lines. 
Furthermore,  a  rhumb  line  on  the  earth  is  represented  by  a 
straight  line  on  the  map. 

*G.  Kremer,  the  latinized  form  of  whose  name  was  Mercator,  com- 
pleted a  map  of  the  world  on  the  plan  here  set  forth  in  1569. 


APPLICATIONS  TO   THE   GEOMETRY  OF  SPACE     333 

We  call  attention  to  the  fact  that  the  above  map  cannot  be 
obtained  by  projecting  the  points  of  the  sphere  on  the  cylinder 
along  a  bundle  of  rays  from  the  centre. 

EXERCISE 

Turn  to  an  atlas  and  test  the  Mercator's  charts  there  found 
by  actual  measurement  and  computation. 


CHAPTER   XVI 

TAYLOR'S  THEOREM  FOR  FUNCTIONS  OF  SEVERAL 
VARIABLES 

1.  The  Law  of  the  Mean.  Let  f(x,  y)  be  a  continuous  func* 
tion  of  the  two  independent  variables  x  and  y,  having  continuous 
first  partial  derivatives.     We  wish  to  obtain  an  expression  for 

f(x0  +  h,y0  +  k) 

analogous  to  the  Law  of  the  Mean  for  functions  of  a  single 
variable,  Chap.  XI,  §  2.  One  such  expression  has  been  found 
in  Chap.  XIV,  §  6;  but  there  is  a  simpler  one.  Form  the 
function  : 

*(*)  =/(**>  +  th,  2/0  +  tic),         0  <  *  <  1, 

where  x0,  y0,  h,  k  are  constants  and  t  alone  varies.     Notice  that 

<D (1)  =/O0  +  h,  2/0  +  k),  ®  (0)  -/(«Vi  2/o). 

If  we  apply  the  Law  of  the  Mean,  p.  230,  Formula  (Af),  to 
<i>  (t),  setting  a  =  0,  6  =  1,  we  get : 

*(l)=fc(O)+l.*'(0),  O<0<1. 

Now  *'(*)  =  hfx  (x0  +  th,  2/0  +  th)  +  kfy (x0  +  th,  y0  +  th). 

Hence  f(%o  +  h,  y0  +  k)  = 

(1)      /(%*  Sfe)  +  hfx  (x0  +  M,  2/o  +  Oh)  +  A;/,  (a*,  +  6h,  y0  +  0fc)> 

where  0  <  0  <  1,  and  this  is  the  form  we  sought  for  the  Law  of 
the  Mean  for  functions  of  two  independent  variables. 
The  extension  to  functions  of  n  >  2  variables  is  obvious. 

334 


FUNCTIONS   OF   SEVERAL  VARIABLES  835 

2.  Taylor's  Theorem.  We  obtain  Taylor's  Theorem  with  the 
Remainder  if  we  write  the  corresponding  theorem  for  $(£): 

*(i)=*(0)+*'(<>)  + ...  +^(w)(0)+     1lv*ln+1)(0), 

and  then  substitute  for  <£  and  its  derivatives  their  values.  Thus 
when  w  =  lwe  get 

(2)  f(x0  +  h,  2/0  +  &)  =f(x0, 2/0)  +  hfm (a%,  2/o) 4-  &/„  (a*>,  2/o) 

+|[tf/>(X,  F)  +  2M-/xy(X,  F)+*«/>(X,  F)], 
where  X  =  #0  +  07*,    F=  y0  -+■  0&,  and  0  <  6  <  1. 

The  student  should  write  out  the  formula  for  the  next  ease, 

71  =  2. 

The  general  term,  3>(n)  (0)/w !,  can  be  expressed  symbolically  as 
nl\_  dx        dy_\  x=x0 

iy=v0 

and  the  remainder  as 

(W  +  1)!|       £#  0v]  x  =  z0  +  *A 

|y  =  yo  +  0* 

The  extension  to  functions  of  n  >  2  variables  is  immediate. 

If  the  remainder  converges  toward  zero  when  n  becomes 
infinite,  we  obtain  an  infinite  series  whose  terms  are  homo- 
geneous polynomials  and  which  converges  toward  the  value  of 
the  function.  If  furthermore  the  series  whose  terms  consist  of 
the  monomials  that  make  up  the  terms  of  the  latter  series  con- 
verges for  all  values  of  h  and  k  within  certain  limits  :  \h\<H9 
\k\<K,  we  say  that  the  function  can  be  developed  in  a  power 
series  in  h  =  x  —  x0  and  k  =  y  —  y0: 

(3)  f(x,  y)=  2 cmn (x  -  x0)m  (y  -  2/o)n, 

or  that  it  can  be  developed  by  Taylor's  Theorem.  A  series  of 
the  form  (3)  is  often  called  a  Taylor's  Series.  But  it  is  not  in 
general  feasible  to  show  that  the  remainder  converges  toward 
zero,  and  so  other  methods  of  analysis  have  to  be  employed  to 
establish  a  Taylor's  development. 


336  CALCULUS 

3.  Maxima  and  Minima.  The  function  f(x,  y)  will  have  a 
maximum  at  the  point  (x0,  y0)  if  the  tangent  plane  of  the 
surface 

at  (x0,  y0)  is  parallel  to  the  x,  y  plane  and  the  surface  lies 
below  this  plane  at  all  other  points  of  the  neighborhood  of 
(x0>  y0,  u0).     Hence  we  see  that  at  (a?0,  y0) 

A  similar  statement  holds  for  a  minimum. 

The  necessary  condition  contained  in  (4)  can  -be  extended 
at  once  to  functions  of  n  >  2  variables.  For,  if  any  one  of  the 
first  partial  derivatives,  du/dx,  for  example,  were  ^0  at 
(#0,  y0,  z0,  •••),  then  the  function  f(x,  y0,  z0,  •••),  a  function  of 
x  alone,  would  be  increasing  as  x  passes  through  the  value  x0, 
or  else  it  would  be  decreasing,  according  to  the  sign  of  du/dx. 

The  conditions  (4)  are  frequently  sufficient  to  determine  a 
maximum  or  a  minimum. 

Example  1.  Given  three  particles  of  masses  mlf  m2,  ra3, 
situated  at  the  points  (xl}  y^,  (x29  y2),  («3,  2/3).  To  find  the 
point  about  which  the  moment  of  inertia  of  these  particles  will 
be  a  minimum. 

Here  it  is  clear  that  for  all  distant  points  of  the  plane  the 
moment  of  inertia  is  large,  becoming  infinite  in  the  infinite 
region  of  the  plane.  Furthermore,  the  moment  of  inertia  is  a 
positive  continuous  function.     Hence  the  surface 

u  =  I=m1  [(«  -xiy  +  (y-yiy]+m2[(x-  x2)2  +  (y  -  y2)2~\ 

+  ra3[(a  -  x3)2  +  (y  -  2/3)2] 

must  have  at  least  one  minimum,  and  at  such  a  point 

Y  =  2  [>!  (oj  —  xY)  +  ra2  (x  —  x2)  +  m3(x  —  a$]  =  0, 
y  =  2[m1(y-y1)+m2(y--y2)  +  m;i(y-y3)']  =  0. 


FUNCTIONS   OF   SEVERAL  VARIABLES  337 

But  these  equations  determine  the  centre  of  gravity  of  the 
particles  and  are  satisfied  by  no  other  point.  Hence  the  centre 
of  gravity  is  the  point  about  which  the  moment  of  inertia  is 
least. 

The  result  is  in  accordance  with  the  general  theorem  of 
Chap.  IX,  §  15,  and  it  holds  for  any  system  of  particles 
whatever. 

Auxiliary  Variables.  As  in  the  case  of  functions  of  a  single 
variable,  so  here  it  frequently  happens  that  it  is  best  to  express 
the  quantity  to  be  made  a  maximum  or  a  minimum  in  terms 
of  more  variables  than  are  necessary,  one  or  more  relations 
existing  between  these  variables.  The  student  must,  therefore, 
in  all  cases  begin  by  considering  how  many  independent  varia- 
bles there  are,  and  then  write  down  all  the  relations  between 
the  letters  that  enter;  and  he  must  make  up  his  mind  as  to 
what  letters  he  will  take  as  independent  variables  before  he 
begins  to  differentiate. 

Example  2.  What  is  the  volume  of  the  greatest  rectangular 
parallelopiped  that  can  be  inscribed  in  the  ellipsoid : 

We  assume  that  the  faces  are  to  be  parallel  to  the  coordinate 
planes  and  thus  obtain  for  the  volume : 

V=  8  xyz. 

But  x,  y,  z  cannot  all  be  chosen  at  pleasure.  They  are  con- 
nected by  the  relation  (5).  So  the  number  of  independent 
variables  is  here  two,  and  we  may  take  them  as  x  and  y.  We 
have,  then : 

£-«»(■+■©-*• 

From  (5)  we  obtai  n : 


338  CALCULUS 

dz  __  _  c2x  dz  _      c2y 

dx         a?z'  dy         b2z 

Now  neither  x  =  0  nor  y  =  0  can  lead  to  a  solution,  and  the 
only  remaining  possibility  is  that 

^  _  y2  _  z2 
a2~62~c2' 

Thus  the  parallelopiped  whose  vertices  lie  at  the  intersections 
of  these  lines  with  the  ellipsoid,  i.e.  on  the  diagonals  of  the 
circumscribed  parallelopiped  x  =  ±  a,  y—±b,  z  =  ±  c,  is  the 
one  required,*  and  its  volume  is 

pr_  Sabc^ 
_3V3* 

EXERCISES 

1.  Required  the  parallelopiped  of  given  volume  and  mini- 
mum surface.  Ans.   A  cube. 

2.  Required  the  parallelopiped  of  given  surface  and  maxi- 
mum volume.  A?is.   A  cube. 

3.  A  tank  in  the  form  of  a  rectangular  parallelopiped,  open 
at  the  top,  is  to  be  built,  and  it  is  to  hold  a  given  amount  of 
water.  Find  what  proportions  it  should  have,  in  order  that 
the  cost  of  lining  it  may  be  as  small  as  possible.  How  many 
independent  variables  are  there  in  this  problem  ? 

Ans.    Length  and  breadth  each  double  the  depth. 

*  The  reasoning,  given  at  length,  is  as  follows.  V  is  a  continuous 
positive  function  of  x  and  y  at  all  such  points  of  the  quadrant  of  the  ellipse 

«2  x  ft*       ' 

for  which  x  >  0,  y  >  0,  and  it  vanishes  on  the  boundary  of  this  region. 
Hence  it  must  have  at  least  one  maximum  inside.  But  we  find  only  one 
point,  as  =  a/V3,  y  =  b/y/B  at  which  V  can  possibly  be  a  maximum. 
Hence,  etc. 


FUNCTIONS  OF  SEVERAL  VARIABLES  339 

4.  Find  the  shortest  distance  between  the  lines 

y  =  2x,  |       y  =  3x  +  7f 

z=5x,  |       z  —  x. 

5.  Show  without  using  the  calculus  that  the  function 

x4  +  y*  +  4:X  —  32  2/  —  7 
has  a  minimum. 

Suggestion.     Use  polar  coordinates. 

6.  Find  the  minimum  in  the  preceding  problem. 

7.  A  hundred  tenement  houses  of  given  cubical  content  are 
to  be  built  in  a  factory  town.  They  are  to  have  a  rectangular 
ground  plan  and  a  gable  roof.  Find  the  dimensions  for  which 
the  area  of  walls  and  roof  will  be  least.* 

8.  A  torpedo  in  the  form  of  a  cylinder  with  equal  conical 
ends  is  to  be  made  out  of  boiler  plates  and  is  just  to  float 
when  loaded.  The  displacement  of  the  torpedo  being  given, 
what  must  be  its  proportions,  that  it  may  carry  the  greatest 
weight  of  dynamite  ? 

Ans.  The  length  of  the  torpedo  must  be  three  times  the 
length  of  the  cylindrical  portion,  and  the  diameter  must  be  V5 
times  the  length  of  the  cylindrical  portion. 

9.  Find  the  point  so  situated  that  the  sum  of  its  distances 
from  the  three  vertices  of  an  acute-angled  triangle  is  a  mini- 
mum. 

Ans.  The  lines  joining  the  point  with  the  vertices  make 
angles  of  120°  with  one  another,  f 

10.  Find  the  most  economical  dimensions  for  a  powder 
house  of  given  cubical  content,  if  it  is  built  in  the  form  of  a 
cylinder  and  the  roof  is  a  cone. 

*  The  problem  is  identical  with  that  of  finding  the  best  shape  for  a 
wall-tent. 

t  For  a  complete  discussion  of  the  problem  for  any  triangle  see  Goursat- 
Hedrick,  Mathematical  Analysis,  vol.  1,  §  62. 


340  CALCULUS 

11.  Find  approximately  the  most  economical  dimensions  for 
a  two-gallon  milk  can.  Assume  the  upper  part  of  the  can  to 
be  a  complete  cone. 

4.  Test  by  the  Derivatives  of  the  Second  Order.  We  proceed 
to  deduce  a  sufficient  condition  for  a  maximum  or  a  minimum 
in  terms  of  the  derivatives  of  the  second  order.  Suppose  the 
necessary  conditions  (4)  are  fulfilled  at  (x0,  y0).  Then  from 
(2)  we  get : 

(6)       f(x0  +  h,  y0  -f-  k)  -f(x0,  y0)  =  i  (Ah2  +  2Bhk  +  Ck2), 

where        A  =fx*  (x0  +  Oh,  y0  +  Bk),    B  =fxy  (x0  +  Oh,  y0  +  $k), 

C=fy*(x0  +  6h,y0  +  dk), 

and  for  a  minimum  the  difference  (6)  must  be  positive  for  all 
points  x=.x0  +  h,  y  =  y0-{-k  near  (x0,  y0)  except  for  this  one 
point,  where  it  vanishes. 

Definite  Quadratic  Forms.  A  homogeneous  polynomial  of 
the  second  degree  in  any  number  of  variables  is  called  a  quad- 
ratic form,*  and  is  said  to  be  definite  if  it  vanishes  only  when 
all  the  variables  vanish.     Thus 

7i2  +  fc2,         2h2  +  3k2  +  5l2 

are  examples  of  definite  quadratic  forms  in  two  and  three 
variables  respectively ; 

h2,  Sh2  +  Ihk  +  2k2  =  (3h  +  k)(h  +  2k), 

regarded  as  quadratic  forms  in  two  variables,  are  not  definite. 
A  definite  quadratic  form  never  changes  sign. 

Theorem.     In  order  that 

U=Ah2  +  2Bhk+Clci, 

*For  some  purposes  it  is  desirable  to  define  an  algebraic  form  merely 
as  a  polynomial.  But  we  are  concerned  here  only  with  homogeneous  poly- 
nomials.   Moreover,  we  exclude  the  case  that  all  the  coefficients  vanish. 


FUNCTIONS  OF  SEVERAL  VARIABLES  341 

where  A,  B,  C  are  independent  of  h  and  k,  be  a  definite  form,  it 
is  necessary  and  sufficient  that 

(7)  B2-AC<0. 

That  this  condition  is  sufficient  is  at  once  evident.  For,  if 
it  is  fulfilled,  surely  neither  A  nor  C  can  vanish,  and  we  can 
write : 

U  =  -[(Ah  +  Bk)2+(AC-  £2)&2]. 
A. 

Hence  U  can  vanish  only  when 

Ah  +  Bk=0      and      fc  =  0, 

i.e.  only  when  h  =  k  =  0,     q.  e.  d. 

We  leave  the  proof  that  the  condition  is  necessary  to  the 
student. 

When  the  condition  (7)  is  fulfilled,  A  and  C  necessarily  have 
the  same  sign,  and  this  is  the  sign  of  U. 

Corollary.  If  A,  B,  C  depend  on  h  and  k  in  any  manner 
whatever,  and  if,  for  a  pair  of  values  (h,  k)  not  both  zero,  the  con- 
dition (7)  is  fulfilled,  then  for  these  values  U  has  the  same  sign  as 
A  and  C. 

Application  to  Maxima  and  Minima.  Returning  now  to  equa- 
tions (6),  let  us  suppose  that 

^8)  \dx~d^)~dx^df<0 

at  (x0,  y0)  and  that  these  derivatives  are  continuous  in  the 
vicinity  of  this  point.  Then  the  relation  (8)  will  hold  for  all 
points   near  (x0,  y0)  and  furthermore,  for  such  points,  both 

—  and  — -^  will  preserve  the  sign  they  have  at  (#0,  y0)-  Hence 
dx2  dy2 

the  right-hand  side  of  (6)  will  vanish  only  at  (x0,  y0),  and  at 

other  points  in  the  neighborhood  will  have  the  sign  common 

to  these  latter  derivatives.     We  are  thus  led  to  the  following : 


342 


CALCULUS 


Sufficient   Condition  for  a  Maximum  or  a  Minimum. 
If  at  the  point  (x0,  y0) 

-v  tiu     A        du 

(a)  —  =  0        —  =  0 


(&> 


dx 

d2u\ 


dy 


/j^Y-  — —  <0 
\dxdy)      dx2  dy2 


and  if  the  derivatives  of  the  second  order  are  continuous  near 
(x0,  y0),  then  u  will  have  a  maximum  at  (x0,  y0)  if 

d2u 


and  a  minimum  there  if 


dx2 


<o, 


g#5 


Conditions  (6)  and  (c)  are  not  necessary,  but  only  sufficient. 
u  may  have  a  maximum  or  a  minimum  even  when  the  sign  of 
inequality  in  (6)  is  replaced  by  the  sign  of  equality.  But  if,  in 
(6),  the  sign  of  inequality  is  reversed,  u  has  neither  a  maximum 
nor  a  minimum. 

When / depends  onw>2  variables,  the  method  of  procedure 
is  similar.  First,  the  algebraic  theorem  about  quadratic  forms 
has  to  be  generalized.     Thus  for  three  variables, 

(9)    U—  aux^  +  a22x22  +  a^xz2  +  2a12x1x2  +  2a13xxx3  +  2a23x2x3, 

and  the  necessary  and  sufficient  condition  that  U  be  a  positive 
definite  quadratic  form  is  that 

an  a12  a13 


(10)  On>0, 


an  a^ 

a21  a22 


>o, 


G&21  ^22  ^23 

a31  a32  a.33 


>o, 


where  av  =  ajt.     This  form  of  statement  suggests  the  general- 
ization for  n  =  n. 

If  U  is  to  be  a  negative  definite  quadratic  form,  the  first, 
third,  fifth,  etc.  inequality  signs  in  (10)  must  be  reversed.  For 
a  proof  by  Gibbs,  arranged  by  Saurel,  cf .  the  Annals  of  Mathe- 
matics, ser.  2,  vol.  4  (1902-03),  p.  62. 


FUNCTIONS  OF  SEVERAL  VARIABLES  343 

The  case  of  implicit  functions,  treated  by  Lagrange's  multi- 
pliers, is  given  in  Goursat-Hedrick,  Mathematical  Analysis, 
vol.  1,  §  61. 

EXERCISES 

1.  Show  that  the  surface 

z  =  xy 

has  neither  a  maximum  nor  a  minimum  at  the  origin. 

2.  Test  the  function 

x?  +  3^  _  2xy  -f  5y2  -  4^ 
for  maxima  and  minima. 

3.  Determine  the  maxima  and  minima  of  the  surface 

x?  +  2  y2  +  3  z2  -  2  xy  -  2yz  =  2. 


CHAPTER  XVII 

ENVELOPES 

1.   Envelope  of  a  Family  of  Curves.     Consider  a  family  of 
circles,  of  equal  radii,  whose  centres  all  lie  on  a  right  line  : 

(1)  (x-ay  +  tf^l, 

where  the  parameter  a  runs  through  all  values.     The  lines 

(2)  2/  =  l         and        y  —  ~  1 

are  touched  by  all  the  curves  of  this  family. 

Again,  let  a  rod  slide  with  one  end  on  the 

floor  and  the  other   touching  a  vertical  wall, 

the  rod  always  remaining  in  the  same  vertical 

plane.     It  is  clear  that  the  rod  in  its  successive 

positions  is  always  tangent  to  a  certain  curve. 

This  curve,  like  the  lines  (2)  in  the  preceding 

Fig.  86  example,  is  called  the  envelope  of  the  family  of 

curves. 

Turning  now  to  the  general  case,  we  see  that  the  family  of 

curves 

(3)  f(x,y,a)  =  0 

may  have  one  or  more  curves  to  which,  as  a  varies,  the  succes- 
sive members  of  the  family  are  tangent.  When  this  is  so,  two 
curves  of  the  family  corresponding  to  values  of  a  differing  but 
slightly  from  each  other  : 

(4)  fix,  y,  «0)  =  0,  fix,  y,  a0  +  A«)  =  0, 

344 


ENVELOPES  345 

will  usually  intersect  near  the  points  of  contact  of  these  curves 
with  the  envelope,  as  is  illustrated  in  the  above  examples.  So 
if  we  determine  the  limiting  position  of  this  point  P  of  inter- 
section of  the  curves  (4),  we  shall  obtain  a  point  of  the  enve- 
lope.    Now  a  third  curve  through  P  is  the  following : 

(5)  0  =f(x,  y,ao  +  Aa)  —f(x,  y,  «b)  =  Aa/a  (x9  y>  Oq  +  0Aa). 

For,  the  coordinates  of  P  satisfy  the  equation  of  this  curve. 
Hence,  allowing  Aa  to  approach  0,  we  get* 

(6)  /*  0,2/,  0=0. 

Thus  the  coordinates  of  a  point  of  the  envelope,  when  one 
exists,  are  seen  to  satisfy  the  simultaneous  equations : 

f        /(®>y>a)=o, 

Conversely,  the  locus  (7)  will  be  tangent  to  each  curve  (3) 
provided  that  df/dx,  df/dy  do  not  both  vanish  along  this  locus. 
To  prove  this,  observe  that  the  slope  of  a  curve  of  the  family 
(3)  is  given  by  the  equation  : 

(8)  ^  +  ^  =  0. 
dx     dy  dx 

In  order  to  find  the  slope  of  the  envelope,  we  may  think  of 
equations  (7)  as  solved  for  x  and  y : 

(9)  x=<j>(a),  2/  =  <K«). 

*  The  reasoning,  in  detail,  is  as  follows.  We  assume  that  the  coordi- 
nates x,  y  of  the  point  P  vary  continuously  as  Aa  approaches  0,  and 
approach  a  definite  limiting  point.  The  coordinates  of  P  satisfy  (5)  and 
hence 

fa  (x,  y,  ao  +  OAa)  =  0. 

Finally,  we  assume  fa  (x,  y,  a)  to  be  ^continuous  function  of  x,  y,  and  a, 
and  so 

Urn  fa (x,  y,  uo  +  db>a)  =fa(x,  y,  «o)  =  0. 


346 


CALCULUS 


Then  the  slope  of  the  envelope  is 

dy  =^'(a)i 
dx      <j>\a) 

Now  take  the  total  differential  of  /  (x,  y,  a) : 

df=dJ-dx  +  d4-dy  +  dif-da. 
ox  cy  Ca 

If  x  and  y  satisfy  (9),  then  df=0,  dx  =  <f>'  (a)  da,  dy  =  if/' (a)  daf 

and  -^  =  0.     Hence 
da 


(10) 


0=dldx  +  lfdy 
ox  cy 


or 


dx      cy  <f>'(a) 


0. 


Thns  (10)  gives  the  same  slope  that  (8)  does,  and  the  envelope 
is  tangent  to  the  family. 

Example  1.  Applying  the  formulas  (7)  to  the  family  of  circles 
(1)  we  get : 

|£==_2(.t-«)  =  0. 
ca 


The  elimination  of  a  between  this  equation  and  (1)  gives 


f 


or 


2/  =  l  and  y  —  —  l. 


Example  2.   To  find  the  envelope  of  the  family  of  ellipses 
whose  axes  coincide  and  whose  areas  are  constant. 
Here, 


Fig.  87 


(a) 
(P) 


"1  "t"  Ti  ~"     > 

a2     62 
Trab  =  k. 


It  is  more  convenient  to  retain  both  param- 
eters, rather  than  to  eliminate,  but  we  must 
be  careful  to  remember  that  only  one  is  inde- 
pendent. If  we  choose  a  as  that  one,  a  =  a,  and  differentiate 
with  respect  to  a,  we  have  ; 


ENVELOPES  347 

a3       b3  da  \  da) 

and  hence 

Between  (a),  (6),  and  (c)  we  can  eliminate  a  and  6  and  thus  get 
a  single  equation  in  #  and  y,  which  will  be  the  equation  of  the 
envelope.  To  do  this,  solve  (a)  and  (c)  for  a2  and  62,  thus 
getting 

a2  =  2aj*,  &2  =  22/2, 

and  then  substitute  the  values  of  a  and  6  from  these  equations 
in  (6): 

±  2-rrXy  =  K, 

a  pair  of  equilateral  hyperbolas. 
The  equations 

x=  ±  a^/2,  y  =  ± bV2, 

combined  with  (b),  give  the  coordinates  of  the  points  of  the 
envelope  in  which  the  particular  ellipse  corresponding  to  that 
pair  of  values  of  a  and  b  is  tangent  to  it.  This  remark  applies 
generally  whenever  the  coordinates  x  and  y  of  a  point  of  the 
envelope  are  obtained  as  functions  of  a. 

EXERCISES 

In  each  of  the  following  questions  draw  a  rough  figure  to  indi- 
cate the  curves  of  the  family  and  the  envelope. 

1.  Find  the  envelope  of  the  family  of  parabolas : 

y2  =  Sax  —  a3. 

2.  Circles  are  drawn  on  the  double  ordinates  of  a  pa- 
rabola as  diameters.  Show  that  their  envelope  is  an  equal 
parabola. 

3.  Show  that  the  envelope  of  all  ellipses  having  coincident 
axes,  the  straight  line  joining  the  extremities  of  the  axes  being 
of  constant  length,  is  a  square. 


348  CALCULUS 

4.  Find  the  envelope  of  straight  lines  drawn  perpendicular 
to  the  normals  of  a  parabola  at  the  points  where  they  cut  the 
axis. 

5.  Show  that  the  envelope  of  the  lines  in  the  second  exam- 
ple of  §  1,  p.  344,  is  an  arc  of  a  four-cusped  hypocycloid. 

6.  The  legs  of  a  variable  right  triangle  lie  along  two  fixed 
lines.  If  the  area  of  the  triangle  remains  constant,  find  the 
envelope  of  the  hypothenuse. 

7.  Find  the  envelope  of  a  circle  which  is  always  tangent  to 
the  axis  of  x  and  always  has  its  centre  on  the  parabola  y  =  x2. 

8.  What  is  the  envelope  of  all  the  chords  of  a  circle  which 
are  of  a  given  length  ? 

9.  Find  the  envelope  of  the  family  of  circles  which  pass 
through  the  origin  and  have  their  centres  on  the  hyperbola 
xy  =  l. 

10.  A  straight  line  moves  in  such  a  way  that  the  sum  of 
its  intercepts  on  two  rectangular  axes  is  constant.  Find  its 
envelope.     Draw  an  accurate  figure. 

11.  The  streams  of  water  in  a  fountain  issue  from  the 
nozzle,  which  is  small,  in  all  directions,  but  with  the  same 
velocity,  vQ .  Show  that  the  form  of  the  fountain  is  approxi- 
mately a  paraboloid  of  revolution. 

2.  Envelope  of  Tangents  and  Normals.  Any  curve  may  be 
regarded  as  the  envelope  of  its  tangents.  Thus  the  equation 
of  the  tangent  to  the  parabola 

(ii) 

at  the  point  (x0,  y0)  is 

y-yo="-(?-x0) 

or 
(12) 


y2  —  2ma 

y0  =  —  (x- 

mx     2/0 

'      2/o       2 

ENVELOPES  349 

Hence  the  envelope  of  the  lines  (12),  where  y0  is  regarded  as  a 
parameter,  must  be  the  parabola  (11),  and  the  student  can 
readily  assure  himself  that  this  is  the  case. 

The  evolute  of  a  curve  was  defined  as  the  locus  of  the 
centres  of  curvature,  and  it  was  shown  that  the  normal  to 
the  curve  is  tangent  to  the  evolute.  Hence  the  evolute  is  the 
envelope  of  the  normals,  and  thus  we  have  a  new  method  for 
determining  the  evolute. 

For  example,  the  equation  of  the  normal  to  the  parabola 

at  the  point  (x0,  y0)  is 

x-x0  +  2x0(y-y0)=0 
or  x-\-2x0y  —  x0  —  2#03  =  0, 

and  we  get  at  once  as  the  envelope  of  this  family  of  lines : 

y  =  SxQ2  +  J,  x  =  —  4a?03, 

or  &-«?-«*- 


EXERCISES 

1.  Obtain  the  equation  of  the  evolute  of  the  ellipse : 

x  ==  a  cos  <£,  y  =  b  sin  <f>, 

as  the  envelope  of  its  normals. 

2.  Obtain  the  evolute  of  the  cycloid  : 

x  =  a(0  —  sin  0),  y  =  a(l  —  cos0). 

3.  Obtain  the  coordinates  (a^,  ?/i)  of  any  point  on  the  en- 
velope of  the  normals  to  the  curve  y  =f(x): 

v — a*, + f  («b)  (y  —  y0 ) = 0, 

and  show  that  the  result  agrees  with  the  formulas  of  Chap. 
VII,  §  3. 


350 


CALCULUS 


Fig.  88 


3.    Caustics.     When  rays  of  light  that  are   nearly  parallel 
fall  on  the  concave  side  of  a  napkin  ring  or  a  water  glass,  a 
portion  of  the  table  cloth  is  illuminated.     Let  us 
determine  the  equation  of  the  boundary. 

Suppose  we  have  a  narrow  semicircular  band, 
on  the  polished  concave  side  of  which  a  bundle 
of  parallel  rays  fall.  The  rays  are  reflected  at  the 
same  angle  with  the  normal  as  the  angle  of  inci- 
dence, and  so  we  wish  to  find  the  envelope  of  the  reflected  rays. 
Take  the  radius  of  the  band  as  1.  Then  the  equa- 
tion of  the  reflected  ray  is 

(13)  y  -  sin  0  =  tan  2  0  (x  -  cos  0). 

To  get  the  envelope  of  the  family,  we  differentiate 
with  respect  to  0  : 

—  cos  0  =  2sec22  0  (x  —  cos  0)  +  tan  2  0  sin  0, 

2<c  ==  2cos  0  —  cos220 cos  0  —  cos 20  sin  20  sin  0 

=  2cos  0  -  cos  20 (cos  20  cos  0  +  sin  20  sin  0) 

=  2cos  0  —  cos  20  cos  0, 

or  :  x  =  i(3cos  0  -  2cos30). 

Substituting  this  value  of  x  in  (13)  we  get : 

y  =  sin30. 

But  these  are  the  equations  of  an  epicycloid  of  two  cusps,  i.e. 
the  one  in  which  a  =  2  b,  b  —  J,  p.  150,  (9). 


Fig.  89 


EXERCISE 

If  the  band  is  a  complete  circle  and  a  point-source  of  light 
is  situated  on  the  circumference,  draw  accurately  a  figure 
showing  the  reflected  rays  and  prove  that  their  envelope  is  a 
cardioid. 


CHAPTER   XVIII 
DOUBLE  INTEGRALS 

1.  Volume  of  Any  Solid.  In  Chap.  IX  we  have  computed 
the  volumes  of  a  number  of  solids  more  or  less  irregular  in 
shape.  It  is  not  difficult  to  generalize  and  obtain  a  method 
for  computing  the  volume  of  any  solid  whatsoever  by  integra- 
tion. A  suggestive  example  is  given  by  a  problem  of  naval 
architecture,  — that  of  determining  the  displacement  of  a  ship. 
Here,  the  plans  of  the  ship,  drawn  on  paper  to  scale,  furnish 
the  areas  of  cross-sections  which  are  near  enough  together  so 
that  a  good  approximation  for  the  volume  of  the  ship  between 
two  successive  cross-sections  may  be  obtained  by  considering 
this  part  of  the  ship  as  a  cylinder  whose  base  is  one  of  the 
cross-sections  and  whose  altitude  is  the  distance  to  the  next 
one.* 

Let  us  now  conceive  a  solid  of  arbitrary  shape.  Assume  a 
line  in  space,  whose  direction  is  taken  at  pleasure,  and  cut  the 
solid  by  a  variable  plane  perpendicular  to  this  line ;  see  Fig.  90. 
Denote  the  distance  of  an  arbitrary  point  on  the  line  from  a 
fixed  point  of  the  line  by  x.  The  area  of  the  cross-section  made 
by  the  above  plane  is  a  function  of  x,  which  we  will  denote  by 
A  (x),  or  simply  A.  Let  the  minimum  x  corresponding  to  one 
of  the  above  planes  be  x  =  a,  the  maximum,  x  =  b.  Divide  the 
interval    from    a    to    b    into    n    equal    parts    by   the    points 

*  It  is  possible  to  approximate  to  the  volume  still  better  by  means  of 
more  elaborate  formulas  (Simpson's  Rule),  but  this  simplest  approxima- 
tion is  more  suggestive  for  our  present  purposes. 

351 


352 


CALCULUS 


x0  =  a,  xlf  •••,  xn  =  b  and  pass  planes  through  these  points 
perpendicular  to  the  line.  Then  the  volume  in  question  is 
given  approximately  by  the  sum  : 

A  (xq)  Ax  -j-  A  .(ajj)  Ax  -\ \-A(xn_1)  Ax, 

and  the  limit  of  this  sum,  when  n  becomes  infinite,  is  exactly 

the  volume  sought : 


a) 


b 
=   CAte. 


Fig.  90 


Example.     To  compute  the  vol- 
ume of  the  ellipsoid : 

a2^b2     c2       ' 
Here,   the   cross-section  made  by 


an  arbitrary  plane  x  =  x'  is  the  ellipse 
b2      c2  a2' 


ir 


+ 


Its  semiaxes  have  respectively  the  lengths 

^F5 


=  i. 


•v-^ 


and  hence  its  area  is,  the  accents  being  suppressed : 
The  volume  Fis,  therefore, 

—a 
,   /  X3  \\a 

\       3aV|_0     * 


DOUBLE   INTEGRALS 


353 


2.  Two  Expressions  for  the  Volume  under  a  Surface ;  First 
Method.  We  turn  now  to  the  problem  of  computing  the 
volume  under  any  surface, 


(2) 


*=/0,  2/)- 


Fig.  91 


Given,  namely,  a  region  S  of  the 

(x,  y) -plane  and  a  function  f(x,  y), 

single    valued     and    continuous 

throughout   S ;    for  the   present 

we  will  assume,  furthermore,  that 

/  is  positive.    Erect  a  cylindrical 

column  on  S  as  base  and  consider 

the  volume  of  the  part  of  this  column  capped  by  the  surface 

(2).     It  is  this  volume  Fthat  we  wish  to  compute. 

Our  first  method  is  that  of  §  1.  We  cut 
the  solid  by  a  plane  x  =  x'  and  compute  the 
area  A  of  this  cross-section.  Now  A  is 
merely  the  area  under  the  curve 


y=r, 


Fig. 


V=T0 
92 


z  =  <f>(y)  =f(x',  y)         (x'}  constant) 
between    the    ordinates    corresponding    to 


the  abscissas  y=Y0  and  y  =  Yx.     Hence 


-/ 


f(x',  y)  dy. 


Dropping  the  accent,  which  has  now 
served  its  purpose,  we  have : 

(3)  A(x)=Jf(x,y)dy, 


where  we  must  remember  that  x  is  constant,  y  being  the  vari- 
able of  integration,  and  that  F0  and  Yx  are  functions  of  x. 

It  remains  only  to  integrate  A  with  respect  to  x  between 
the  limits  x  =  a  and  x  =  b,  where  a  is  the  smallest  abscissa 


354  CALCULUS 

that  any  point  in  S  has,  and   b   is  the  largest.     We  thus 
obtain :  h 


-J 


A  (x)  dx. 


This  last  integral  is  commonly  written  in  either  of  the  forms  :  * 

b  Yi  b     Yt 

I  dx  I  f(x,  y)  dy        or  /    /  f(x,  y)  dy  dx. 

a  to  a     Yq 

It  is  called  the  iterated  integral  of  f(x,  y)  (not  the  double 
integral;  the  latter  will  be  explained  later),  since  it  is  the 
result  of  two  ordinary  integrations  performed  in  succession. 

Instead  of  integrating  first  with  regard  to  y  and  then  with 
regard  to  x,  we  might  have  reversed  the  order,  integrating  first 
with  regard  to  x.     We  should  thus  obtain  the  formula : 


0  Xt 

V  =  fdyjf(x, 


y)dx. 

For  example,  let  us  compute  the  volume  cut  off  from  the 

paraboloid : 

^      x2     y2 

z  =  l 2- 

4      9 

by  the  (x,  2/)-plane.  Since  the  surface  is  obviously  symmetric 
with  respect  both  to  the  (x,  z)  and  the  (y,  z)  planes,  it  is  suffi- 
cient to  compute  the  part  of  the  volume  that  lies  in  the  first 
octant,  and  then  multiply  the  result  by  4.     To  get  A  we  have 

*  Another  form  sometimes  employed  is  to  be  avoided,  namely : 

b    Yt 

§{[f(x,y)dxdy. 

viation 


The  second  form  given  in  the  text  is  to  be  thought  of  as  an  abbre 

for 

b      r, 


J|j/(z,  2/)#}<fc. 


DOUBLE  INTEGRALS 


355 


to  hold  x  fast,  i.e.  to  cut  the  solid  by  the  plane  x  =  x',  and 
compute  the  area  of  the  section.  This  is  the  area  under  the 
curve 

the  limits  of  integration  being  determined  as  follows.  The 
(x,  y)  plane,  whose  equation  is  z  =  0,  cuts  the  surface  in  .the 
ellipse 

0  =  1 


x2     y2 


and  the  region  S  is  the  part  of  this 
ellipse  lying  in  the  first  quadrant. 
The  segment  of  the  line  x  =  x'  which 
lies  within  S  has  for  its  minimum 
ordinate  F0  =  0,  for  its  maximum  Ylf 
where 


Fig.  94 


~'2  XT2 

0  =  1-  —  -4l 


Yi=4V4-  x'2. 


Thus 


^/(1-t-!M1-t>-!!>{'-t"--S}>-. 

0 

=  i(4-z'2)V4-o;'2. 

Hence,  dropping  the  accent,  we  get : 

A  =  $(±-x2)i 

Finally,   integrating   A  from   the  smallest  x  in  S  to  the 
largest,  we  have  (see  Tables,  No.  137)  : 

2 

1    C(±-x2)§dx  = 

iL(4-^  +  6«V4=^+24sin-1|~|2  =  ^, 
16|_  2J0       4 

and  so  the  total  volume  is  37r  =  9.42. 


356  CALCULUS 


EXERCISES 


1.  A  round  hole  of  radius  unity  is  bored  through  the  solid 
just  considered,  the  axis  of  the  hole  being  the  axis  of  z.  Find 
the  volume  removed. 

2.  Compute  the  volume  of  a  cylindrical  column  standing  on 
the  area  common  to  the  two  parabolas 

x  =  y2,  y  =  x> 

as  base  and  cut  off  by  the  surface 

z  =  12  +  y-x2. 

3.  Work  each  of  the  foregoing  examples,  integrating  first 
with  regard  to  x  and  then  with  regard  to  y. 

3.  Continuation.  Second  Method.  Another  way  of  finding 
the  above  volume  is  as  follows.  Divide  the  region  S  up  into 
small  pieces,  called  elements  of  area,  of  arbitrary  shape,  and 
denote  the  area  of  any  one  of  them  by  b.Sk.  Let  (xk,  yk)  be  an 
arbitrary  point  of  the  kth.  element.  Construct  a  cylinder 
on  this  element  as  base  and  of  height  f(xk,  yk)  ;  see  Fig.  102. 
The  volume  of  this  column  is 

Consider  now  the  totality  of  such  columns.  They  form  a 
solid  whose  volume, 

(4)  2/(*»y,)AS„ 

differs  only  slightly  from  the  volume  T^we  wish  to  compute. 
As  n  grows  larger  and  larger,  the  maximum  diameter  of  each 
of  the  elementary  areas  approaching  0  as  its  limit,  it  is  clear 
that  the  limit  of  (4)  is  V: 

(5)  V=limV  f(xk,yk)ASk. 

This  is  the  second  expression  for  the  volume  we  set  out  to 
obtain. 


DOUBLE  INTEGRALS 


357 


Z 


I 


Fig.  95 


We  remark  that  it  is  not  important  that  the  elementary- 
areas  just  fill  out  the  region  S.  Thus  we  might  divide  the 
plane  by  parallels  to  the  coordinate 
axes  into  rectangles  whose  sides  are 
of  length  Ax  and  Ay,  and  then  take 
as  the  elementary  areas  (a)  all  the 
rectangles  that  lie  wholly  within  S  j 
or  (6)  all  those  just  mentioned  and 
in  addition  such  as  contain  at  least 
one  point  of  the  boundary  of  S  in 
their  interior  or  on  their  boundary  ;  or  (c)  any  set  intermediate 
between  (a)  and  (&).  In  each  case  the  sum  (4)  would  clearly 
have  as  its  limit  the  volume  V. 

4.  The  Fundamental  Theorem  of  the  Integral  Calculus.  Just 
as  in  Chap.  IX,  §  2,  we  equated  the  two  expressions  for  the 
area  under  a  curve  to  each  other  and  thus  obtained  an  analyti- 
cal theorem  regarding  limits,  so  here  we  equate  the  two  expres- 
sions just  found  for  the  volume  under  a  surface  and  thereby 
deduce  a  corresponding  theorem  for  functions  of  two  inde- 
pendent variables. 

Fundamental  Theorem  of  the  Integral  Calculus.  Let 
f(x,  y)  be  a  continuous  function  of  x  and  y  throughout  a  region  S 
of  the  (x,  y)-plane.  Divide  this  region  up  into  n  pieces  of  area 
ASq,  ASu  •••,  A$n_!  and  form  the  sum: 

f(x„  y0)  ASo+ZO-!,  ft)  A£i+  '••  +/<X-i,  *-4)AflU, 

where  (xk,  yk)  is  any  point  of  the  k-th  elementary  area.  If  n  now 
be  allowed  to  increase  without  limit,  the  maximum  diameter  of 
each  of  the  elements  of  area  approaching  0  as  its  limit,  this  sum 
will  approach  a  limit  which  is  given  by  the  formula : 

by"  &  x" 

J  dx  f  f(x,  y)  dy        or  /  dy  J  f  (x,  y)  dx, 


where  the  limits  of  s  Uegration  are  determined  as  described  in  §  2. 


358  CALCULUS 

Expressed  as  a  formula  the  theorem  is  as  follows  : 

b  y"  p  x" 

(6)  Km  ^  f(xk,  yk)  ±Sk  =fdxff(x,  y)dy  =  Jdy  f*f(x,  y) dx. 

*  =  0  ay'  ax' 

Definition  of  the  Double  Integral.  The  limit  that 
stands  in  the  first  member  of  (6)  is  called  the  double  integral  of 
the  function  /  taken  over  the  region  S,  and  is  written  as 
follows  : 

(7)  lim  ^  f(xk,  yk) ASk  =JffdS. 

It  is  independent  of  the  particular  system  of  coordinates  used, 
and  applies  equally  well,  whether  cartesian  or  polar  coordinates 
are  employed.     The  iterated  integral,  on  the  other  hand,  has 
been  obtained  at  present  only  for  cartesian  coordinates. 
The  double  integral  is  also  written  in  the  form : 

I    jfdxdy         or  /    IfrdrdO, 

the  latter  form  referring  to  polar  coordinates  (cf.  §  7). 

The  Fundamental  Theorem  can  now  be  written  as  follows : 


&  y" 

(6')  JJfdS=JdxJf(x, 


y)dy, 


with  a  similar  formula  when  the  first  integration  is  performed 

with  respect  to  x. 

We  have  hitherto  assumed  that  the  boundary  of  S  is  cut  by 
a  parallel  to  the  axis  of  y  at  most  in  two 
points.  If  this  is  not  the  case,  there  is 
still  no  difficulty  in  the  definition  of  the 
double  integral.  For  the  purpose  of 
evaluating  the  same,  however,  by  means 
Fia.  96  of  the  iterated   integral,  S  may  be  di- 

vided up  into  regions,  for  each  of  which 

the  above  is  true  (see  Fig.  96),  and  then,  inasmuch  as  the  double 

integral  extended  over  all  S  is  evidently  equal  to  the  sum  of 


DOUBLE  INTEGRALS  359 

the  double  integrals  of  the  same  function  extended  over  the 
different  divisions  of  S,  it  is  sufficient  to  compute  the  double 
integral  for  each  of  these  divisions  by  means  of  (6). 

We  have  further  assumed  that  the  function  /  is  positive  in 
JS.  If  it  were  negative,  the  same  reasoning  would  still  hold, 
only  both  expressions  for  V  would  yield 
the  negative  value  of  the  volume.  They 
would,  therefore,  still  be  equal  to  each 
other.  If,  finally,  /  changes  sign  in  S, 
divide  S  up  into  regions  in  which  S  is 


Fig.  97 


positive  and  those  in  which  it  is  nega- 
tive.    The  Fundamental  Theorem  holds  for  each  region  by 
itself,  and  so  it  holds  for  the  combined  region. 


EXERCISE 


Show  that  the  abscissa  of  the  centre  of  gravity  of  a  homo- 
geneous plane  area  is  given  by  the  formula : 

IJxdS 

5.  Moments  of  Inertia.  Consider  the  moment  of  inertia  of 
a  plane  lamina  of  variable  density  p  about  a  point  0  in  its 
plane.  In  accordance  with  Chap.  IX, '  §  14,  we  divide  the 
lamina  up  into  small  pieces,  of  area  ASk  and  of  mass  AMk ,  and 
form  the  sum: 

*=o 
where  rk  is  the  distance  of  a  point  (xk,  yk)  of  the  fcth  elemen- 
tary area  from  0.     We  can  write  the  mass  AMk  as  the  product 
of  the  corresponding  area  &Sk  by  the  average  density  of  this 

n-1 

Hence  I—  lim  x  p*?*2  A£fc. 


360  CALCULUS 


If  the  first  factor,  pki  in  each  term  is  the  value  of  p  in  the  par- 
ticular point  (xk,  yk),  then  the  limit  of  this  sum  is  by  defini- 
tion the  double  integral 

If,  however,  this  is  not  the  case,  we  need  only  to  apply  Du- 
hamel's  Theorem,  setting 

where  pk  is  the  value  of  p  in  (xk,  yk). 

Then  lim&  =  l, 

and  hence  in  all  cases 

(8)  I=JJPT*dS. 

S 

Example  1.     The  density  of  a  "rectangle  is  proportional  to 
the  square  of  the  distance  from  one  corner.     Find  its  moment 
of  inertia  about  that  corner. 
Here,  p  =  Xr2, 

and  hence  I=X  I    I  r4dS; 

s 

a  b 

J   J7AdS  =  fdx  l(x4  +  2x2y2  +  if)dy  =  la5b  +  %a3b3  +  iab5; 

I=^-(9a4  +  10a2b2  +  9b*). 
4o 

The  mass  of  any  lamina  is  easily  seen  to  be 

(9)  M=ffPdS. 

s 
In  the  present  case,  therefore, 


DOUBLE   INTEGRALS  361 

Hence  Ja8  ^rf  +  lOaW  +  W 

15(a2  +  62) 

It  is  sometimes  more  convenient  to  use  the  formulation  of 
the  moment  of  inertia  as  a  double  integral,  even  when  the  den- 
sity of  the  lamina  is  constant,  e.g. : 


Example  2.     To  find  the  moment  of  inertia  of  a  triangular 
lamina  of  constant  density  about  a  vertex. 
Here, 

'"  S 


,  f  fr*dS;  ^-H    x 

J  J  01  x  =  h 

3  h         y»  Fig.  98 

C  Cr*dS  =   fdm  /V  +  y2)  dy, 

S  0  y' 

<f  wm  I'm,  y"  =  l"x. 

h*      Mh2 
.'.    I=Pil"-V  +  \(r*-Vz)]j  =  ^(2>  +  V2  +  VV'  +1"*). 

EXERCISES 

1.  Determine  by  double  integration  the  moment  of  inertia  of 
a  right  triangle  of  constant  density  about  the  vertex  of  the 
right  angle.  A         M(a?  +  b2) 

6 

2.  Compute  the  moment  of  inertia  about  the  focus  of  the 
segment  of  a  parabola  cut  off  by  the  latus  rectum. 

3.  Show  that  the  moment  of  inertia  of  a  lamina  about  the 
axis  of  y  is  „  „ 

1=  J    I  px2dS. 

4.  Find  the  moment  of  inertia  about  the  axis  of  y  of  a 
uniform  lamina  bounded  by  the  parabola  y2  =  4ax,  the  line 
x  +  y  =Sa,  and  the  axis  of  x.  Work  the  problem  both  ways, 
integrating  first  with  regard  to  x,  then  with  regard  to  y\  and 
then  in  the  opposite  order.  .  »    46  pa4 


362  CALCULUS 

6.  Theorems  of  Pappus.  Theorem  I.  If  a  closed  curve  rotate 
about  an  external  axis  lying  in  its  plane,  the  volume  of  the  ring 
thus  generated  is  the  same  as  that  of  a  cylinder  whose  base  is  the 
region  S  enclosed  by  the  curve  and  whose  altitude  is  the  distance 
through  which  the  centre  of  gravity  of  S  has  travelled: 

(10)  .    V=27rh-A, 

where  h  denotes  the  distance  of  the  centre  of  gravity  of  S  from 
the  axis,  and  A,  the  area  of  S. 

We  will  confine  ourselves  to  the  case  that  the  boundary 
curve  is  met  at  most  in  two  points  by  a  parallel  to  the  axis 
of  rotation,  which  we  will  take  as  the  axis  of  ordinates. 
Divide  the  area  into  strips  of  breadth  Ax  by  parallels  to  the 
axis  of  y,  and  approximate  to  the  volume  generated  by  the  fcth 
strip  by  means  of  the  volume  generated  by  a  rectangle  with 
the  left-hand  boundary  of  this  strip  for  one  of  its  sides  and 
with  base  Ax*  This  latter  volume  can  be  computed  at  once 
as  the  difference  between  two  cylinders  of  revolution,  and  is 

*4+i(y'!e-yfk)  -  WW-y'k)  =27rxM-yd  ^ +»(j£-i©  ^% 

y'z=<f>(x)   being  the  equation   of   the   lower    boundary,   and 
y"  =f(x)  that  of  the  upper  one.     Hence 

F=limV  \27rxk(y:-y'k)Ax  +  'n'(y:-y'k)Axil 

This  last  expression  can  be  simplified  by  DuhameFs  Theorem, 
and  thus 

n-l  h 

F=lim  y\27rxk(yZ-yk)Ax  =  2TT  fx(y"-y')dx. 

a     ■ 

Recalling  the  result  of  Ex.  4,  p.  174,  we  see  that  the  value 
of  this  integral  is  xA  =  hA,  and  this  completes  the  proof. 

If  the  curve  rotates  only  through  an  angle  ©  instead  of 
completely  round  the  axis,  we  have  merely  to  replace  2ir  by  ©. 

*  The  student  should  draw  the  requisite  figure. 


DOUBLE  INTEGRALS  363 

Finally,  the  form  of  the  proof  is  somewhat  simplified  by- 
means  of  double  integrals,  the  above  restriction  on  the  boundary, 
as  well  as  the  use  of  Duhamel's  Theorem,  being  then  unneces- 
sary.    We  have  at  once : 

C  CxdS. 
F=lim^27rxA^=27r  C  CxdS,  ^=~§~A 

Theorem  II.  If  a  plane  curve,  closed  or  not  closed,  rotate 
about  an  axis  not  cutting  it  and  lying  in  its  plane,  the  area  of  the 
surface  thus  generated  is  the  same  as  that  part  of  the  cylindrical 
surface  having  the  given  curve  as  generatrix,  which  lies  between 
two  parallel  planes  whose  distance  apart  is  the  distance  traversed 
by  the  centre  of  gravity  of  the  given  curve : 

S  =  2tt7i'1      or       ®h-L 

The  proof  is  similar  to  that  of  the  first  theorem,  and  is  left 
as  an  exercise  for  the  student. 

7.  Polar  Coordinates.  We  have  computed  the  volume  V 
under  the  surface  z=f(x,  y)  by  iterated  integration,  using 
cartesian  coordinates.  Let  us  now  compute  the  same  volume, 
using  polar  coordinates.  To  do  this  we  ^ 
divide  the  solid  up  into  thin  wedge- 
shaped  slabs  (the  slab  not  extending  in 
general  clear  to  the  edge  of  the  wedge) 
by  means  of  n  equally  spaced  planes 
through  the  axis  of  z :  6  =  00  =  a,  Bx,  •••, 
0n  =  /?,  and  approximate  to  the  volume 

of  the  A;-th  slab,  AVk,  as  follows.  Let  Ak  be  the  area  of  the 
section  of  the  plane  0  ==  0k  with  the  solid,  and  let  this  section 
rotate  about  the  axis  of  z  through  the  angle  A0.  Then,  by  the 
first  theorem  of  Pappus,  §  6,  the  volume  generated  is  A0  •  hkAk, 
and  the  sum  of  such  volumes, 

k=0 


364  CALCULUS 

is  a  good  approximation  for  V.  In  fact,  when  we  visualize  the 
totality  of  these  pieces,  we  see  that  the  volume  of  the  solid 
thus  obtained  approaches  Fas  its  limit,  when  71  =  00.     Hence 

8 

(11)  V=  Hm  V  hk  Ak  A0=    ChAdB. 

—  JTo  s7 

Furthermore,  let  us  consider  the  product  hA  corresponding 
to  the  cross-section  made  by  an  arbitrary  plane  6  =  6'.  Writing 
the  equation  of  the  surface  in  the  form 

z=f(;x,y)  =  F(r,B) 

and  recalling  the  general  formula  for  the  centre  of  gravity : 

b 
I  xydx 

we  have  here  to  set 

x—r,      x  —  h,      y  =  z  =  F (r,  &),      a  =  r'}      b  =  r", 
and  we  thus  obtain : 

r" 

hA=  frF(r,  6')  dr. 

r' 

Substituting  this   last  expression  in  (11),  we  get  the  final 

formula : 

i 


F=      d6     rF(r,  6)dr, 
and  hence  the 
Theorem  : 

6  r" 

(12)  C  Cf(t,  6)dS=  jd$  frF(r,  6) dr. 


DOUBLE   INTEGRALS 


365 


The  first  integration  is  performed  on 
the  supposition  that  6  is  held  fast  and 
that  r  varies  from  the  smallest  value 
r',  which  it  has  in  JS  corresponding  to 
the  given  value  of  0  to  the  largest 
value,  r". 


Fig.  100 


TJie  Inverse  Order  of  Integration.     If   instead  of  using  the 
planes  $  =  00,  $lt  --,0n  we  had  divided  the  solid  up  by  the  cylin- 


ders r-. 
result : 


r«  =  a,  r. 


*n  =  b,  we  should  have  been  led  to  the 


U  V 

(13)  f  fF(r,8)dS=  fdr  frF(r,  0)d$. 


\&*~-._  / 


Here,  the  first  integration  is  performed 
on  the  supposition  that  r  is  held  fast 
and  that  0  varies  from  the  smallest 
value,  $',  which  it  has  in  S  correspond- 
ing to  the  given  value  of  r  to  the  largest 

value,  0". 
Fig.  101  ' 

Example.    To  find  the  moment  of  inertia  of  a  uniform  circu- 


lar disc  about  its  centre.     Here 


2ir  a 

=  P  f  fr'dS^p  fdO  fr3dr  =  P- 


and  hence 


/=  Ma2/2. 

This  problem  we  have  solved  before  by  single  integration. 
The  solution  by  double  integration  is  simpler  in  form,  though 
in  substance  the  two  solutions  are  closely  related. 


EXERCISES 

1.  The  density  of  a  circular  disc  is  proportional  to  the  dis- 
tance from  the  centre.  Find  the  radius  of  gyration  of  the  disc 
about  its  centre.  Ans.   aV|. 


366  CALCULUS 

2.  Determine  the  moment  of  inertia  about  the  focus  of  the 
segment  of  the  parabola : 

r=       w 
1  —  cos  0 

bounded  by  the  latus  rectum. 

3.  The  density  of  a  square  lamina  is  proportional  to  the 
distance  from  one  corner.  Find  its  moment  of  inertia  about 
this  corner. 

4.  Find  the  moment  of  inertia  about  the  origin  of  the  part 
of  the  first  quadrant  bounded  by  two  successive  coils  of  the 
equiangular  spiral 

r  —  e9, 

the  inner  boundary  going  through  the  point  6  =  0,  r  =  1. 

5.  Find  the  moment  of  inertia  of  the  lemniscate : 

r2  =  a2  cos  2  6, 
about  the  point  r  =  0. 

6.  Show  that  the  abscissa  of  the  centre  or*  gravity  of  any 
plane  area  is  given  by  the  formula: 


SJ> 


pxdS 


M 

7.  Find  the  centre  of  gravity  of  the  lemniscate  of  question  5. 

8.  Show  that  the  area  of  any  plane  region  S  is  expressed  by 
the  integrals : 

A=   I    fdxdy=  f   IrdrdB. 

s  s 

9.  Find  the  area  bounded  by  the  curve 

0  =  sin  r 

and  the  portion  of  the  axis  of  x  between  the  origin  and  the 
point  x  =  7r.  Ans.  it. 


DOUBLE   INTEGRALS 


367 


8.  Areas  of  Surfaces.  We  have  determined  the  area  under 
a  plane  curve  and  the  lateral  area  of  a  surface  of  revolution  by 
means  of  simple  integrals.  The  general  problem  of  finding  the 
area  of  any  curved  surface  is  solved  by  double  integration. 

Let  the  equation  of  the  surface  be 

***/(*,  y) 

and  let  the  projection  on  the  x,  y  plane  of  the  part  @  of  this 
surface  whose  area  A  is  to  be  computed,  be  the  region  S. 
Divide  8  up  into  elementary  areas  and  erect  on  the  perimeter 
of  each  as  generatrix  a  cylindrical  surface.  By  means  of  these 
cylinders  the  surface  @  is  divided  into  elementary  pieces,  of  area 
&Ak,  (k  =  0,  1,  •••,  Ti  —  1),  and  we  next  consider  how  we  may 
approximate  to  these  partial  areas.  Evidently  this  may  be 
done  by  constructing  the  tangent  plane  at  a  point  (xk,  yk,  zk)  of 
the  Avfch  elementary  area  and  computing  the  area  cut  out  of  this 
plane  by  the  cylinder  in  question.  Now  the  orthogonal  cross- 
section  of  this  cylinder  is  of  area  b£ki  and  hence  the  oblique 
section  will  have  the  area 


A/^secy,, 

where  yk  is  the  angle  between  the 
planes,  or  between  their  normals. 
The  desired  approximation  is 
thus  seen  to  be 


n — i 

^  &Sk  sec  yk, 


Fig.  102 


and  consequently  A  is  equal  to  the  limit  of  this  sum,  or 


*  It  is  a  fundamental  principle  of  elementary  geometry  to  refer  all  geo- 
metrical truth  back  directly  to  the  definitions  and  axioms.  What  are  the 
axioms  on  which  this  formula  depends  ?  The  answer  is  :  The  formula 
itself  is  an  axiom.  The  justification  for  this  axiom  is  the  same  as  for  any 
other  physical  law,  namely,  that  the  physical  science,  here  geometry,  built 
on  it  is  in  accord  with  experience. 


368 


CALCULUS 


(14) 


A  =  I    j  sec  y  dS. 


The  angle  y  is  the  angle  between  the  normal  to  the  surface 
and  the  axis  of  z.     Hence  by  Chap.  XV,  §  1 : 


(15) 


dx2     dy2 

If  the  equation  of  the  surface  is  written  in  the  form 

F(x,y,z)  =  0, 
we  have 

(16)  sec 


^IH!)*+© 


dF 
cz 


Example.  Two  equal  cylinders  of  revolution  are  tangent  to 
each  other  externally  along  a  diameter  of  a  sphere,  whose  radius 
is  double  that  of  the  cylinders.  Find  the  area  of  the  surface 
of  the  sphere  interior  to  the  cylinders. 

It  is  sufficient  to  compute  the  area  in  the  first  octant  and 
multiply  the  result  by  8.  We  have  to  extend  the  integral  (14) 
over  the  region  S  indicated  in  Fig.  104.     Here, 


x2  -+-  y2  +  z2  —  a2, 


and  by  (16) 


sec  v  =  -  = 


Va2  —  r2 


r2  =  x2  +  y2. 


Fig.  103 


Since  the  integrand,  sec  y,  depends  in 
£  *  a  simple  way  on  r,  it  will  probably  be 
well    to   use  polar   coordinates   in   the 
iterated  integral.     We  have,  then: 


M-I/1-'-- /-/^S 


DOUBLE  INTEGRALS 


369 


J 


a  cos  9 

*    ardr 


=  —  a  Va2  —  r2 


^/cf  —  r2 


a  cos  & 

=  a2(l-sin0), 


2 

'•  I  ^  =  a" T(l  -  sin  0)  d£  =  a2(|  -1 V 


yl  =  4  7ra2-8a2. 


Fia.  104 


Objection  may  be  raised  to  the  foregoing  solution  on  the 
ground  that  the  integrand,  sec  y  =  a/ Va2  —  r2,  does  not  remain 
finite  throughout  S,  but  becomes  infinite  at  the  point  0  =  0, 
r  =  a.  We  may  avoid  this  difficulty  by  computing  first  only  so 
much  of  the  area  as  lies  over  the  angle  a  ;<  0  <i  -rr/2,  where  the 
positive  quantity  a  is  chosen  arbitrarily  smalL  The  value  of 
this  area  is 


«./(!- 


sin0)<20  =  a2 


a  — cos 


■> 


and  its  limit,  when  a  approaches  0,  is 


EXERCISES 

1.  A  cylinder  is  constructed  on  a  single  loop  of  the  curve 
r  =  a  cos  n6  as  generatrix,  its  elements  being  perpendicular  to 
the  plane  of  this  curve.  Determine  the  area  of  the  portion  of 
the  surface  of  the  sphere  a?  +  y2  +  z2  =  2  az  which  the  cylinder 
intercepts.  j  2  (tt  —  2)  a2 


2.    Compute  the  moment  of  inertia  about  the  axis  of  z  of  the 
surface  whose  area  was  determined  above  in  the  text. 


3.    A  square  hole  is  cut  through  a  sphere,  the  axis  of  the  hole 
2b 


370  CALCULUS 

coinciding  with  a  diameter  of  the  sphere.     Find  the  area  of  the 

.surface  removed.      .  Aa  ,    .  _*       b  Q   2  .     .     b2 

Ans.     loaosm  l  —  —  8 or  sin  1— -. 

Va2-62  a  -  &2 

4.  Determine  the  area  of  the  surface 

z  =  xy 

included  within  the  cylinder 

X2  -f-  y2  =  a2. 

5.  M  cylindrical  surface  is  erected  on  the  curve  r  =  0  as 
generatrix,  the  elements  being  perpendicular  to  the  plane  of 
this  curv«.     Find  the  area  of  the  portion  of  the  surface 

z  —  xy 
which  is  bounded  by  the  y,  z  plane  and  so  much  of  the  cylindri- 
cal surface  as  corresponds  to  0  ^  6  <^  tt/2. 

9.  Cylindrical  Surfaces.  If  the  surface  @  is  a  cylinder,  the 
area  can  be  expressed  explicitly  as  a  simple  integral.  Let  the 
elements  of  the  cylinder  be  parallel  to  the  axis  of  y.  The  equa- 
tion of  the  surface  then  becomes : 

Hence        A  =  f  f  secydS  =   I  dx  I  VI +f(x)2  dy, 

S  ay1 

b 

(i7)  a  =  fvr+TW  (y"  -  y') dx- 


EXERCISES 

1.  Two  cylinders  of  revolution,  of  equal  radii,  intersect,  their 
axes  cutting  each  other  at  right  angles.  Show  that  the  total 
area  of  the  surface  of  the  solid  included  within  these  cylinders 
is  16  a2. 

2.  Obtain  formula  (17)  directly,  without  the  use  of  double 
integrals. 


DOUBLE  INTEGRALS 


371 


3.  Write  out  formula  (17)  when  the  elements  of  the  cylinder 
are  perpendicular  (a)  to  the  x,  y  plane ;  (b)  to  the  y,  z  plane. 

4.  Show  that  the  lateral  area  of  that  part  of  either  of  the 
cylinders  discussed  in  the  example  of  §  8  which  is  contained 

In  the  sphere  is  4  a2. 

5.  The  area  of  a  region  S  of  the  x,  y  plane  may  be  written 
in  the  form : 

A=  f  CdS  =   f(y"-y')dx  =  f(x"-x')dy. 

S  a  a 

By  means  of  the  last  formula  compute  the  area  of  the  region 
common  to  the  circle  and  the  parabola : 


arJ  +  2/2  =  16a2, 


2/2  =  6  ax. 


6.   Deduce  from  formula  (14)  the  formula  of  Chap.  IX,  §  8, 
for  the  area  of  a  surface  of  revolution : 


A  =  2tt  Cy^/l+f'ixydx. 


10.  Analytical  Proof  of  the  Fundamental  Theorem.    Carte- 
sian Coordinates.     In  the  sum : 


(is)         SJ/to.y-)^*, 

*=0 

whose  limit  is  the  double  integral 
(19)  fffd8> 


:g: 


xtxin 
Fig.  105 


we  may  choose  as  elementary  areas  rectangles  with  sides  Ax, 
Ay,  thus  making  ASk  =  AxAy,  and  then  add  all  those  terms 
together  which  correspond  to  rectangles  lying  in  a  column 
parallel  to  the  axis  of  y.  This  partial  sum  can  be  represented 
as  follows : 


372  CALCULUS 

where  we  have  assigned  new  indices,  i  and  j,  to  the  coordinates 
of  the  point  (xk}  yk),  and  where  furthermore  we  have  chosen 
the  points  (xk,  yk)  of  this  column  so  that  they  all  have  the 
same  abscissa,  x(. 

If,  now,  holding  xi  and  Ax  fast,  we  allow  q  to  increase  with- 
out limit,  Ay  approaching  0  as  its  limit,  we  have 

n 

H 

(20)  Ax lim  §  f(xt>  y3)  Ay  =  Ax  Cf(xi9  y)dy. 

?=ooi=b  J, 

Next,  we  add  all  the  limits  of  these  columns  together: 

-i  ^ 

^Ax  J  f(xi,y)dy, 

y'i 

and  allow  p  to  increase  without  limit,  Aaj  approaching  0.  This 
gives 

y'i  b       >/" 

lim^  Ax  j  f(xi,y)dy=  f  dx  jf(x,y)dy, 
Pmm  i=°       y'i  a         y' 

i.e.  the  iterated  integral  of  the  Fundamental  Theorem. 

This  method  of  deduction  is  less  rigorous  than  the  former 
one,  for  we  have  not  proven  that  we  get  the  same  result  when 
we  take  the  limit  by  columns  and  then  take  the  limit  of  the 
sum  of  the  columns,  as  when  we  allow  all  the  ASks  to  approach 
0  simultaneously  in  the  manner  prescribed  in  the  definition  of 
the  double  integral.*     It  is  nevertheless  useful  as  giving  us 

*  For  a  complete  analytical  treatment  of  the  subject  of  this  paragraph 
along  the  lines  here  indicated,  which  in  point  of  elegance  and  rigor  leaves 
nothing  to  be  desired,  see  Goursat-Hedrick,  Mathematical  Analysis, 
Chap.  VI. 


DOUBLE  INTEGRALS  373 

additional  insight  into  the  structure  of  the  iterated  integral, 
for  it  enables  us  to  think  of  the  first  integration  as  correspond- 
ing to  a  summation  of  the  elements  in  (18)  by  columns,  and  of 
^che  second  integration  as  corresponding  to  the  summation  of 
these  columns.  Moreover,  when  we  come  to  polar  coordinates 
in  the  next  paragraph,  it  helps  to  explain  and  make  evident 
the  limits  of  integration. 

11.  Continuation;  Polar  Coordinates.  Let  the  region  S 
be  divided  up  into  elementary  areas  by  the  circles  r  =  rif 
r»+i — ?*»  =  Ar,  and  the  straight  lines  0  =  0j}  6j+1  —  0i  =  A0.     Then 

ASk  =  rkkr  A0  +  \  Ar2  A0, 

and  hence,  in  taking  the  limit  of  the  sum  (18),  &Sk  may,  by 
DuhamePs  Theorem,  be  replaced  by  r*ArA0.     Writing 

f(x,y)  =  F(r,0) 
we  have,  therefore, 

f  ffdS  =  lim  V  F(rk,  0k)  rkAr  Ad. 

In  order  to  evaluate  this  latter  limit,  we  may  replace  (rk,  6k) 
by  (rt,  Oj)  and,  holding  6i  fast,  add  together  those  terms  that 
correspond  to  elementary  areas  lying  in  the  angle  between  the 
rays  9  =0j  and  0  =  0J+1,  thus  getting : 

AO^Ffa,  Oj)r{Ar. 
The  limit  of  this  sum,  as  p  =  oo,  is 

Ad  I  F(r,  6j)rdr. 

5 

Next,  add  all  the  limits  thus  obtained  for  the  successive 
elementary  angles  together  and  take  the  limit  of  this  sum. 
We  thus  get 


Fig.  106 


374  CALCULUS 

lim  ^  AS  Cf(v,  6j)rdr=  CdO  /V(r,  0)rdr, 

i.e.  the  first  iterated  integral,  (12),  of  §  7. 

If  on  the  other  hand  we  hold  rt  fast  and  add  the  terms 
that  correspond  to  elementary  areas  lying  in  the  circular  ring 
bounded  by  the  radii  r  ■—  r<  and  r  =  ri+1)  we  get 

and  the  limit  of  this  sum,  when  q  =  oo,  is 

Ar  /V(r0  0)  r<d0  =  r{  Ar  |F(f0  0)  d0. 
Fig.  107    '  e't  e'. 

Adding  all  these  latter  limits  together  and  taking  the  limit 
of  this  sum,  we  have : 

Oi  b  6" 

lim]T  rt-Ar  f  F(r{,  0)d6=  f  rdr  f  F(r,  0)d$i 

P~°°i=0         e'i  «  r 

t.e.  the  second  iterated  integral,  (13),  of  §  7. 

12.  Surface  Integrals.  The  extension  of  the  conception  of 
the  double  integral  from  a  plane  region  JS  to  a  curved  surface 
@  is  immediate.  Let  a  function  /  be  given,  defined  at  each 
point  of  @,  and  let  it  be  continuous  over  @.  Let  @  be  divided 
up  into  a  large  number  of  small  areas,  —  elementary  areas, — 
A®k,  and  let  fk  be  the  value  of  /  at  an  arbitrary  point  of  A@A. 
Form  the  sum : 


X  /**«*• 


*  =  0 


The  limit  of  this  sum  when  n  grows  larger  and  larger  is  the 
surface  integral  off  over  the  region  © : 


\im^fkA®k=JJfd®. 


DOUBLE   INTEGRALS  375 


EXERCISE 

Show  that  the  volume  of  a  closed  surface  is  given  by  the 
surface  integral : 

V=-  f   I  r  cos  (j>d<5, 

where  r  denotes  the  distance  of  a  variable  point  P  of  the  sur- 
face from  a  fixed  point  0  of  space  and  <f>  is  the  angle  that  the 
outer  normal  of  the  surface  at  P  makes  with  the  line  OP 
produced. 

EXERCISES 

1.   Find  the  volume   cut   out  of  the  first   octant  by  the 

cylinders 

z  =  1  —  x2,  x  =  l  —  y2. 


2.   Compute  the  value  of  the  integral : 

ex2+**'dS, 


Ans.   |f. 


//• 


s 
extended  over  the  interior  of  the  circle 

x*  +  y2  =  l.  Ans.  5.40. 

3.   Evaluate 


//' 


(x*-3ay)dS, 


where  S  is  a  square  with  its  vertices  on  the  coordinate  axes, 
the  length  of  its  diagonal  being  2a.  Ans.   \aA. 

4.   Express  as  an  iterated  integral  in  polar  coordinates  the 
double  integral 

'  rfdS, 


> 


extended  over  a  right  triangle  having  an  acute  angle  in  the 
pole.     Give  both  orders  of  integration. 


376  CALCULUS 

5.  Express  the  iterated  integral 

2         2a  cos  9 

fdd  ifrdr 

~  2 

as  a  double  integral,  and  state  over  what  region  the  latter  is 
extended. 

6.  The  same  for 

n 

2  b  esc  9 

(a)  jdoCfrdr; 

2a        V^ay 
(P) 


idy  jfdx. 


7.  Change   the   order   of  integration  in  the  following  in- 
tegrals : 

i         i 

(a)  jdx  Jf(x,y)dy, 

a         y+a 

(b)  Idy  jf(x,y)dx. 

8.  The  density  of  a  square  lamina  is  proportional  io  the 
distance  from  one  corner.     Determine  the  mass  of  the  lamina. 

Arts.    .765  Aa3. 

9.  Find  the  centre  of  gravity  of  the  lamina   in   the  pre- 
ceding question.     A^    -_-_ar7V2-2+31og(l+_V2)l 

8[V2  +  log(l  +  V2)] 

10.  Two  circles  are  tangent  to  each  other  internally.  De- 
termine the  moment  of  inertia  of  the  region  between  them 
about  the  point  of  tangency. 


DOUBLE   INTEGRALS  377 

li.  Find  the  attraction  of  a  uniform  circular  diso  on  a 
particle  situated  in  a  line  perpendicular  to  the  plane  of  the 
disc  at  its  centre. 

12.  Solve  the  same  problem  for  a  rectangular  disc. 

Ans.    K^tan-i ab  . 

13.  Determine  the  attraction  of  a  uniform  rectangle  on  an 
exterior  particle  situated  in  a  parallel  to  two  of  its  sides,  pass- 
ing through  its  centre. 


Ans.   K^log 
lab 


'h±a  t  b  +  Vjli -  a)2  +  b2 
h-a     &  +  V(7i  +  a)2+b2- 


14.  The  intensity  of  light  issuing  from  a  point  source  is 
inversely  proportional  to  the  square  of  the  distance  from  the 
source.  Formulate  as  an  integral  the  total  illumination  of  a 
plane  region  by  an  arc  light  exterior  to  the  plane. 

15.  Compute  the  illumination  in  the  foregoing  question  on 
the  interior  of  the  curve 

r2  =  l-0\ 

the  light  being  situated  in  the  perpendicular  to  the  plane  of 
the  curve  at  r  =  0.  Ans.    2\(l  —  h  cot-1  h). 

16.  One  loop  of  the  curve 

r^  a3  cos  30 

is  immersed  in  a  liquid,  the  pole  being  at  the  surface  and  the 
initial  line  vertical  and  directed  downward.  Find  the  pressure 
on  the  surface.  *        wa3  V3 

8      ' 

17.  One  loop  of  the  lemniscate 

r2  =  a2cos20 

is  immersed  as  the  loop  of  the  curve  in  the  preceding  question. 
Find  the  centre  of  pressure. 

Ans.   Distance  below  the  surface  =  a  V2  ( h  -  V 

\37r     4y 


378  CALCULUS 

x18.   Formulate  the  volume  of   a  solid  of   revolution  as  a 
double  integral. 

19.  The  curve 

cos0  =  3-3r  +  r2 

rotates  about  the  initial  line.     Find  the  volume  of  the  solid 
generated.  .        23  x 

~30" 

20.  Find  the  volume  cut  from  a  circular  cylinder  whose  axis 
is  parallel  to  the  axis  of  z9  by  the  x,  y  plane  and  the  surface 

xy  an  az. 

Assume  that  the  cylinder  does  not  cut  the  coordinate  axes. 

Ans.   =**£. 
a 

21.  A  cone  of  revolution  has  its  vertex  in  the  surface  of  a 
sphere,  its  axis  coinciding  with  a  diameter.  Find  the  volume 
common  to  the  two  surfaces.  Ans.   f  7ra3(l  —  cos4 a). 

22.  Determine  the  volume  of  an  anchor  ring. 

23.  Determine  the  area  of  the  surface  of  an  anchor  ring. 

24.  Find  the  moment  of  inertia  of  an  anchor  ring  about  its 
axis-  Ans.   M@£  +  »\ 

25.  Find  the  area  of  that  part  of  the  surface 

v 
2  a*  tann- 
ic 

which  lies  in  the  first  octant  below  the  plane  z  =  ir/2  and 
within  the  cylinder  x2  4-  y1  —  1. 

26.  Obtain  a  formula  for  the  centre  of  gravity  of  a  curved 
surface  of  variable  density. 

27.  Obtain  a  formula  for  the  components  of  the  attraction 
which  a  surface  of  constant  or  of  variable  density  exerts  on  a 
particle  of  matter  not  lying  in  the  surface. 


DOUBLE  INTEGRALS  379 

Hence  show  that  the  force  with  which  a  homogeneous  piece 
of  the  surface  of  a  sphere  lying  wholly  in  one  hemisphere  and 
symmetrical  with  reference  to  the  diameter  perpendicular  to 
the  base  of  the  hemisphere  attracts  a  particle  situated  at  the 
centre  of  the  sphere  is  proportional  to  the  projection  of  the 
piece  on  the  base. 

28.  Find  the  moment  of  inertia  about  the  origin  of  the 
portion  of  the  first  quadrant  bounded  by  the  curve 

(x  +  l)(2/  +  l)=4, 

correct  to  three  significant  figures. 

29.  Find  the  volume  of  a  column  capped  by  the  surface 

z  =  xy, 

the  base  of  the  column  being  the  portion  of  the  first  quadrant 
in  the  x,  y  plane  which  lies  between  two  successive  coils  of  the 
logarithmic  spiral : 

r  =  ae0. 

Ans.     ~(e^-l)  (e2M-l). 

30.  Find  the  abscissa  of  the  centre  of  gravity  of  the  above 
.column. 

31.  A  square  hole  2b  on  a  side  is  bored  through  a  cylinder 
of  radius  a,  the  axis  of  the  hole  intersecting  the  axis  of  the 
cylinder  at  right  angles.     Find  the  volume  of  the  chips  cut  out. 


Ans.     4  b2  V  a2  —  ¥  -f  4  a2  b  sin"1-- 

Of- 

32.  A  square  hole  26  on  a  side  is  bored  through  a  sphere  of 
radius  a,  the  axis  of  the  hole  going  through  the  centre  of  the 
sphere.     Find  the  volume  of  the  chips  cut  out. 


Ans.    2-, 


■a2  -  Sa^sm-1 a  -*  tan-1    ,     b      -1 

L         V2(a2-62)     «  Va2-2&2J 


CHAPTER   XIX 

TRIPLE  INTEGRALS 

1.  Definition  of  the  Triple  Integral.  Let  a  function  of  three 
independent  variables,/ (x,y, z),  be  given,  continuous  throughout 
a  region  V  of  three  dimensional  space.  Let  this  region  be 
divided  in  any  manner  into  small  pieces,  of  vohrYne  AVk,  and 
let  (xkf  yk,  zk)  be  an  arbitrary  point  of  the  ft-th  piece.  Form 
the  product  f(xk,  yk,zk)  AVk  and  add  all  these  products  together: 

(1)  §/(*,*,  %)AF* 

fcsmO 

When  n  is  made  to  grow  larger  and  larger  without  limit, 
the  greatest  diameter  of  each  of  the  elementary  volumes 
approaching  0  as  its  limit,  the  sum  (1)  approaches  a  limit, 
and  this  limit  is  defined  as  the  triple  or  volume  integral  of  the 
function  /throughout  the  region  V: 

(2)  lim^f(xk,  yk,  zk)AVk  =  JJJfdV. 

It  is  not  essential  that  the  totality  of  the  elementary  volumes 
should  just  fill  out  the  region  V.  We  might,  for  example, 
divide  space  up  into  small  rectangular  parallelopipeds,  the 
lengths  of  whose  edges  are  A#,  Ay,  Az,  and  consider  such  as 
are  interior  to  V,  or  such  as  have  at  least  one  point  of  Fin 
their  interior  or  on  their  boundary. 

The  integral  is  also  written  as  follows : 

380 


TRIPLE  INTEGRALS  381 

/   /    //0>  y,z)dxdydz. 

V 

The  proof  involved  in  the  above  definition,  that  the  sum  (1) 
actually  approaches  a  limit,  has  to  be  given  along  different 
lines  for  triple  integrals,  from  what  was  possible  in  the  case 
of  double  integrals.     There,  we  were  able  to  represent  the  sum 

n — 1 

A  =  0 

by  a  variable  volume  which  obviously  approached  a  fixed 
volume  as  its  limit.  Here,  we  should  need  a  four  dimensional 
space  in  which  to  represent  geometrically  the  sum  (1).  It 
is  necessary,  therefore,  to  fall  back  on  an  analytical  proof. 
Such  a  proof  will  be  found  in  Goursat-Hedrick,  Mathematical 
Analysis,  Vol.  1,  Chap.  VII.  The  proofs  of  this  and  the  later 
theorems  of  this  chapter  belong  properly  to  a  later  stage  of 
analysis.  The  theorems  themselves,  however,  are  easily  in- 
telligible from  their  analogy  with  the  corresponding  theorems 
for  double  integrals,  and  it  is  our  purpose  here  to  state  them 
and  to  explain  their  uses. 

EXERCISES 
1.   Show  that  the  mass  of  a  body,  of  variable  density  p,  is 

PdV, 


and  that 


f-///' 


,JIIpXdV  fffpXdV 

Jjfav  m 

I=ffPdV> 

where  r  denotes  the  distance  of  a  variable  point  from  the  axis. 


382  CALCULUS 

2.  Formulate  as  a  triple  integral  the  attraction  of  a  body- 
on  a  particle  exterior  to  it. 

2.  The  Iterated  Integral.  In  order  to  compute  the  value  of 
the  volume  integral  denned  in  §  1  we  introduce  an  iterated 
integral  The  method  is  that  of  Chap.  XVIII,  §§  10, 11.  Let 
the  region  V  be  divided  up  by  planes  parallel  to  the  coordinate 
planes  into  rectangular  parallelopipeds  whose  edges  are  of 
lengths  Ax,  Ay,  Az,  and  let  us  take  as  our  elements  of  volume 
these  little  solids.  Then  AVk  =  Ax  Ay  Az,  and  the  sum  (1) 
becomes 

n— 1 

We  will  select  from  this  sum  the  terms  that  correspond  to 
elements  situated  in  a  column  parallel  to  the  axis  of  z  and  add 
them  together,  see  Fig.  108 : 

AxAy^f(x{,yJ9  z{)Az9 

1=0 

where  we  have  assigned  new  indices,  i,  j,  and  I,  to  the  co- 
ordinates of  the  point  (xk,  yk,  zk)  and  where  furthermore  we 
have  chosen  the  points  (xk,  yk,  zk)  of  this  column  so  thSt  they 
all  lie  in  the  line  x  =  xi9y  =  yj.  If,  now,  still  holding  xt9  yj9  Ax, 
and  A.v  fast,  we  allow  s  to  increase  without  limit,  Az  approach- 
ing 0,  we  have 

z" 

«— i  /» 

Ax  Aylim  Vf(xif  yjf  zt)  Az  =  Ax  Ay  I  /(«„  ys,  z)dz, 

z 

where  z'  is  the  smallest  ordinate  of  the  points  of  V  on  the  line 
a5  =  a7»>  y  =  Vj)  and  z"  is  the  largest,  —  we  assume  for  simplicity 
that  the  surface  of  V  is  met  by  a  parallel  to  any  one  of  the 
coordinate  axes  which  traverses  the  interior  of  V  in  two  points. 
Next,  we  add  all  the  limits  of  these  columns  together : 


TRIPLE   INTEGRALS 


383 


where  we  have  set 


s> 


f(*,y>*)dz  =  Q(x,y), 


and  take  the  limit  of  this 
sum.  The  region  JS  of  the 
x,  y  plane  over  which  this 
summation  is  extended  con- 
sists of  the  projections  of  V/ 
the   points   of    V   on    that  Fig.  108 

plane,  and  hence  the  limit  of  this  sum  is  the  double  integral 
of  <J>  (x,  y),  extended  over  S : 

b  y" 

lim  V  $  (xi >  Vj)  &x&y=  I    I  <&dS  =  l  dx  I  ®(x,  y)  dy. 

S  ay' 

We  are  thus  led  to  the  final  result : 

Fundamental  Theorem  of  the  Integral  Calculus: 


w 


ffffdV=1ffdSff(x' y' z)dz 

V  8  z' 

b  y"  z" 

=  jdx  \dy  jf(x,y,z)dz. 


Another  notation  for  the  iterated  integral  is  as  follows : 
f  f   lf(x,y,z)dzdydx. 

a      y'    z' 

Any  other  choice  of  the  orders  of  integration  is  equally- 
allowable. 

An  example  or  two  will  serve  to  illustrate  the  process. 

Example.  Find  the  moment  of  inertia  of  a  tetrahedron 
whose  face  angles  at  a  vertex  O  are  all  right  angles,  about  an 
edge  adjacent  to  0. 


384 


CALCULUS 


Take  0  as  the  origin  of  coordinates  and  the  three  adjacent 
edges  as  the  axes.     Then 

/=  p  fff(*?  +  f)  dV=  pfdxfdyf(x>  +  y*)  dz, 

where  the  limits  of  integration  are  as  follows.  First,  the  limit 
z'  =  0  and  the  limit  z"  =  Z  is  the  maximum  ordinate  in  V  cor- 
responding to  an  arbitrary  pair  of  values  x,  y ;  i.e.  the  ordinate 

of  a  point  in  the  oblique  face  of  the 

tetrahedron : 

a     b     c 


F».  109  Hen0e      Z=C' 

and  the  result  of  the  first  integration  is : 


a     by 


<$>  (x,  y)  =  /V  +  f)  dz  =  (x2  +  2/2)  8 


[*H 


y+  i 


Next,  this  latter  function  must  be  integrated  over  the  sur- 
face S  consisting  of  a  triangle  bounded  by  the  positive  axes  of 
x  and  y,  and  the  line 

a     o 
This  double  integral  may  be  computed  by  iterated  integration, 
the  limits  of  integration  for  y  being  y'  =  0  and 


and  those  for  x,  0  and  a.     The  remainder  of  the  computation 
is,  therefore,  as  follows : 


TRIPLE  INTEGRALS  385 


a  r         z 

JdxH{ 


+  y*)dz  =  c^(a*  +  b2); 

M(a2  +  b2) 
10 


The  student  can  verify  the  answer  by  slicing  the  tetrahe- 
dron up  by  planes  parallel  to  the  x,  y  plane  and  employing  the 
result  of  Ex.  1  at  the  end  of  §  5  in  Chap.  XVIII. 

EXERCISES 

1.  Find  the  centre  of  gravity  of  the  above  tetrahedron. 

2.  Determine  the  moment  of  inertia  of  a  rectangular  paral- 
lelopiped  about  an  axis  passing  through  its  centre  and  parallel 
to  four  of  its  edges. 

3.  A  square  column  has  for  its  upper  base  a  plane  inclined 
to  the  horizon  at  an  angle  of  45°  and  cutting  off  equal  inter- 
cepts on  two  opposite  edges.  How  far  is  the  centre  of  gravity 
of  the  column  from  the  axis  ?  *  a?_ 

3h' 
3.   Continuation ;  Polar  Coordinates.   In  space  there  are  two 
systems  of  polar  coordinates  in  common  use,  namely,  spherical 
coordinates  and  cylindrical  coordinates. 

Spherical  Coordinates.  Let  P,  with  the  cartesian  coordinates 
x,  y,  z,  be  any  point  of  space.  Its  spherical  coordinates  are 
defined  as  indicated  in  the  figure.  If  we  think  of  P  as  a  point 
of  a  sphere  with  its  centre  at  0  and  of  radius  r,  then  0  is  the 
longitude  and   <f>  is  the  colatitude  of  P.  z 

We  have 

ic  =  rsin  <£cos#,    . 

y  —  rsin<f>  sin0, 

z  —  r  cos  <£.  Fig.  110 

We  propose  the  problem  of  computing  the  volume  integral 

(5)  |*£<|  /(**<  «m  **) ^Vk  =  jJJfdV 

2c 


386 


CALCULUS 


by  means  of  iterated  integration  in  spherical  coordinates.  For 
this  purpose  we  will  divide  the  region  Tup  into  elementary 
volumes  as  follows.  Construct  (a)  a  set  of  spheres  with  0  as 
their  common  centre,  r  =  rii  their  radii  increasing  by  Ar; 
(b)  a  set  of  half-planes  0  =  0jf  the  angle  between  two  successive 
planes  being  A0;  and  lastly  (c)  a  set  of  cones  <f>=<f>iy  their 
semi-vertical  angle  increasing  by  A<£  :  <f>l+1  —  <f>t  —  A<£.  The 
element  of  volume  thus  obtained  is  indicated  in  Fig.  111.  The 
lengths  of  the  three  edges  that  meet  at  right  angles  at  P  are 
Ar,  rA<f>,  r  sin  <£A0,  and  hence  this  volume  A V  differs  from 

the  volume  of  a  rectangular  parallel- 
opiped  with  the  edges  just  named : 

(6)  r2sin<£ArA0A<£ 

by  an  infinitesimal  of  higher  order : 

AF 


lim 


=  1. 


r2  sin  <j>  Ar  A0  A<£ 

It   follows,  then,   from   DuhamePs 
Theorem  that  in  the  limit  of  the 
sum  (5)  we  may  replace  AT*  by  the  infinitesimal  (6).    If  we  set 


Fig.  ill 


f(x,y,z)  =  F(r,6,<t>), 


we  have 


J  J  j  fdV=  lim 2J  F(r„  6k,  <f>k)rk2  sin  <f>k  Ar  A0  A<£. 


Can  we  evaluate  this  last  limit  by  iterated  integration  ? 
It  is  easy  to  see  that  we  can.  For  the  sum  is  of  the  type  of 
the  sum  (3),  and  hence  the  method  of  §  2  is  applicable.  Fol- 
lowing that  method,  let  us  select,  for  example,  those  terms  for 
which  6  and  <£  have  a  constant  value,  and  add  them  together : 


p-i 


A0A*2JF(ro  $J>  *i)tf  simfcAr, 


where  $j  and  <£,  are  constant.     They  correspond  to  elementary 
volumes  lying  in  a  row  bounded  by  the  planes  0  =  8j  and 


TRIPLE  INTEGRALS  387 

0  =  $j+lf  and  by  the  cones  cf>  =  fa  and  <f>  =  <f>l+1.  Now  allow  p  to 
increase  without  limit,  Ar  approaching  0.  This  gives,  as  the 
limit  of  the  above  sum, 


A0A<£sin<fo  /  r2F(r,  0j}  <f>t)dr, 


where  r '  is  the  distance  of  the  nearest  point  of  V  to  0  on  the 
line  0  =  0j,  <f>  =  <f>l,  and  r",  that  of  the  farthest.  We  assume 
for  simplicity  that  the  surface  of  V  is  met  by  any  one  of  the 
lines : 


6  =  const.,  <f>  =  const.,  r  =  const., 

<f>  =  const.,  r  —  const.,  0  —  const., 


which  traverses  the  interior  of  V,  in  two  points. 
Next,  we  add  all  the  limits  thus  obtained  together 

where  we  have  set 


r' 


and  take  the  limit  of  this  sum.  If  we  interpret  8  and  <f>  as  the 
coordinates  of  a  point  on  the  surface  of  a  sphere  r  =  const, 
(say,  r  =  1),  then  the  region  S  over  which  the  above  sum  is  to 
be  extended  consists  of  those  points  in  which  radii  vectores 
drawn  to  points  of  V  pierce  the  surface  of  this  sphere.  Hence 
the  limit  of  this  sum  is  the  double  integral  of  ^>  (6,  <f>),  extended 
over  S : 

lim2j*(^,^)AdA*=  f  f*(6>  <t>)dS 

s 

/5  <t>"  r" 

=   I  dO  f  sin  <f>d<t>  f  F(r,  6,  <f>)r>dr. 
We  are  thus  led  to  the  following  result: 


CALCUIXS 


::   :lf  ::rf- 
to#;  (6) with  respect 


I7«r    ili 


JJJp*fim+drmd+     and        /"/'/*-- 


w 

*  <m  A 

m   Cd$  fd+  fr*  *in*+ cos  $dr  =  ir(^4~<0. 

Check.  When  a  =  .4,  x  =  J  .4 ;  when  a  =  0,  £  =  f  a. 

The  student  may  solve  the  same  problem,  taking  the  axis  of 
symmetry  as  the  axis  of  z  and  computing  z. 

ndrical  Coordinates.    The  cylindrical 

coordinates  of  a  point  are  defined  as  in  the 

x    accompanying  figure.     They  are  a  combi- 

/         nation  of  polar  coordinates  in  the  a,  y  plane 

and  the  cartesian  z. 


TRIPLE  INTEGRALS  389 

x  =  r  cos  $j        y=r  sin  0,        z  =  z. 

The  element  of  volume  is  shown  in  Fig.  113.     The  lengths  of 
the  edges  adjacent  to  P, —  they  meet  at  right  angles  there, — 
are :   Ar,  r  A0,   Az.     Hence   the   volume   AF  of   the  element 
differs  from  r  Ar  A0  Az  by  an  infinitesimal 
of  higher  order,  and  we  have : 

rArA0Az  °J^» t=-r> 


From    Duhamel's   Theorem   it   follows,    '  "J 

then,  that  in  taking  the   limit  of  the 

sum  (1),  A7i  may  be  replaced  by  rtAr  A0Az,  and  so,  setting 

S (x,y,z)  =  F(r,0,z), 


we 


obtain:         /   /   I  fdV=  lim  ^  F(rk,  0k,  zk)rkArA$\z. 


This  last  limit  can  be  computed  by  iterated  integration  in 
a  manner  precisely  similar  to  that  set  forth  in  the  case  of 
spherical  coordinates.     We  thus  obtain : 

(8)  fff/dV=fd2fd6f/rdr> 

together  with  similar  formulas  yielded  by  adopting  a  different 
order  of  integration. 

The  above  volume  integral  and  the  iterated  integral  are 
also  written  in  the  forms: 

J   J   ffrdrdOdz      and      C  j   ffrdrdBdz. 

r 

Example.  To  find  the  attraction  of  a  cylindrical  bar  on  a 
particle  of  unit  mass  situated  in  its  axis. 

The  magnitude  of  the  attraction  is  evidently* 

*  The  unit  of  force  is  here  taken  as  the  gravitational  unit. 


390 


CALCULUS 


SSf^r- 


Here 

r2  =  r2  +  ^2, 

Hence 


cos^=-  = 


2tt         h+l 


o  ;*  o 

a 


^ 


vV  +  z2  o 


Fig.  114 


=  1- 


h+l  a  h+l 

fdzf  ***  =i-  r  *d* 


Va2  +  z2' 


-  /  _  Va2  +  (ft+  If  +  Va2  +  ft5; 
.'.      A  =  2ttP[1  +  VaF+W -  Va2  +  (ft  +  r)2] . 

EXERCISES 

1.  Determine  the  attraction  of  a  straight  pipe  on  a  particle 
situated  in  its  axis. 

2.  Find  the  force  with  which  a  cone  of  revolution  attracts  a 
particle  at  its  vertex.  Ans.     2irph  (1  —  cos  a). 

3.  Show  that  the  force  with  which  a  piece  of  a  spherical 
shell  cut  out  by  a  cone  of  revolution  with  its  vertex  at  the 
centre  0  attracts  a  particle  at  0  depends,  for  a  given  cone, 
only  on  the  thickness  of  the  shell. 

4.  Prove  the  preceding  theorem  for  any  cone. 

4.  Line  Integrals.  Line  integrals  present  themselves  in 
such  physical  problems  as  that  of  finding  the  work  done  by  a 
variable  force  when  the  point  of  application  describes  a  curve. 


TRIPLE  INTEGRALS  391 

Let  a  plane  curve  C: 

y=f(x)        or        F(x,y)  =  0, 

be  given.     Its  coordinates  can  always  be  expressed  as  func- 
tions of  the  arc  s,  measured  from  an  arbitrary  point.     Thus  in 

the  case  of  the  circle 

x2  +  y2  =  a2 
we  can  write 

s  .    s 

x  =  a  cos  - ,  y  =  a  sin-, 

a'  9  a' 

where  s  is  measured  from  the  point  (a,  0).     We  will  think  of 
the  equation  of  the  carve  C,  therefore,  as  expressed  in  the 
form : 
(1)  x=<f>(s),  y  =  $(8). 

Consider  next  a  function  F(s)  defined  at  each  point  of  the 
curve.  It  may  be  given  as  a  function  both  of  the  coordinates 
x,  y  of  a  variable  point  P  of  the  plane  and  of  the  arc  s: 
f(x,  y,  s).  But  in  the  latter  case  P  is  to  lie  on  C,  and  so 
x  and  y  have  the  values  given  by  (1),  /  (x,  y,  s)  thus  becoming 
a  function  of  s  alone : 

fix,  y,  s)=f[<f>(s),  $(8),  s]  =  F(s). 

We  will  now  divide  the  arc  up  into  n  equal  parts  by  the  points 
s0  =  0,  slf  •  •  •,  sn_1}  sn  =  I  and  form  the  sum : 

fc  =  0 

The  limit  of  this  sum  as  n  becomes  infinite  is 

i 


/■ 


F(s)ds, 


and  is  called  the  line  integral  of  the  function  F(s)  or  f(x,  y,  s) 
taken  along  C.     Other  notations  for  this  integral  are 

I  fix,  y,  s)  ds      and        /  f(x,  y,  s)  ds, 


392  CALCULUS 

where  (xQ,  y0)  and  (xl}  yx)  are  the  coordinates  of  the  extremities 
of  the  arc  C. 

Geometrically  the  line  integral  admits  of  a  simple  interpreta- 
tion. Let  a  cylinder  be  constructed  on  C  as  generatrix,  its 
elements  being  perpendicular  to  the  x,  y  plane,  and  let  the 
values  of  the  function  F(s)  be  laid  off  along  the  elements  of 
this  cylinder.  Then  the  area  of  the  cylinder  bounded  by  this 
curve  and  the  generatrix  represents  the  line  integral  in  question. 
As  an  example  of  a  line  integral,  suppose  a  point  moves  in  a 
field  of  force.  Let  the  magnitude  of  the  force  be  g  and  let  the 
force  make  an  angle  6  with  the  tangent  to  C  drawn  in  the  direc- 
tion of  the  motion.  Then  the  compo- 
nent of  the  force  along  the  curve  is 
gcos#,  and  the  work  done  by  the 
force  is  / 


(2)  W=  I  g  cos  dds 


Fig.  115 


/f 


A  Second  Form  of  the  Line  Integral.    A  second  form  in  which 
line  integrals  appear  is  the  following : 

I  Pdx+Qdy, 

the  meaning  of  the  integral  being  this.  Two  functions  P(x,  y), 
Q  (x,  y)  of  the  independent  variables  x,  y  are  given,  the  curve  is 
divided  as  before,  and  the  sum 

n-l 

(3)  .    2  lp(x*>  2/*)  Aa*  +  Q  (xk,  Vk)  A^.] 

fc  =  0 

is  formed,  Axk  denoting  the  difference  xk.+1  —  xk,  and  similarly 
for  Ay*.     The  limit  of  this  sum  is  the  limit  in  question. 

To  evaluate  the  limit,  we  may  write  the  summand  in  the 
form: 

(P(x>,  y.)  ^+  Q(x„,  yk)  £»W 


TRIPLE  INTEGRALS 


393 


Now 


v       Ax 

lim  —  =  cost, 

a«=o  As 


lim  — *  =  sin  t, 

As=0  AS 


and  hence  by  Duhamel's  Theorem  the  limit  of  (3)  and  the  limit 
of  the  sum 

n— 1 

2}  [P(a?*,  yk)co8Tk+-Q(xk,  yk)smTk']  As 

4=0 

are  the  same.     But  the  limit  of  the  latter  sum  is 

f [P(a>,  y)  cost  +  Q(x9  y)  sin r]  ds  =  /Ti^  +  ^ffW 

where  the  x  and  2/  in  the  integrands  are  given  by  (1). 

As  an  example  of  the  second  form  of  line  integral  consider 
again  a  field  of  force,  the  components  of  the  force  along  the 
axes  being  denoted  at  each  point  by  X,  Y.  Then  the  work 
done  by  the  force  when  the  point  of  application  describes  the 
curve  C  is 

(4)  W=  fxdx+Ydy. 

The  relation  between  formulas  (2)  and  (4)  for  the  work  be- 
comes clear  when  we  consider  the  special  case  that  the  point 
of  application  P  moves  in  a  right  line,  the  force  not  changing 
in  magnitude  or  direction.  One  expression  for  the  work, — 
that  corresponding  to  (2),  —  is 

W=(gcos0)Z. 

On  the  other  hand,  the  work 
done  by  the  component  X  is 

(Xcos  t)  I  =  X  (xj  —  x0), 

and  that  done  by  Yf 


Fig.  116 


(Fsinr)  J  =^-2/0). 
Hence  we  ought  to  have : 

$lcos6=X(x1-x0)+  Y(yl-y0). 


394  CALCULUS 

That  this  is  in  fact  a  true  relation  is  readily  seen.     For  the 
component  of  gf  along  the  line  P  describes,  namely 

PM=%  cos  e, 

is  equal  to  the  sum  of  the  components  of  X  and  Y,  namely, 

PN=Xcosr      and      NM  =  Fsinr. 

But  cost  =  ^— ^2,  sinr  =  ^^. 

(  i 

Hence  gcos0  =  X^^+F  2l=-&, 

i  I 

and  thus  the  above  relation  is  seen  to  be  true. 

When  the  force  changes  and  the  path  is  a  curve,  we  still 
have 

gcos0  =  XcosT+rsinT=X—  +  F^, 

ds  ds 

and  hence  g  cos  Qds  =  Xdx  4-  Fdy. 

&pace  Curves.     Both  line  integrals  admit  of  immediate  ex- 
tension to  space  curves  C : 

X=<f>(s),  y  =  $(s),  «  =  «(*), 

the  first  integral  giving 

i  1  1 

//(*>  y>  *$  s)ds=  //[>(«),  if,(s)y  »(«),  s]ds  =  I F(s)ds, 

and  the  second, 


Prite  +  Qdy  +  Rdz. 

Thus  in  the  case  of  a  field  of  force  we  should  have  for  the 
work  : 

TT=  jXdx+Ydy  +  Zdz. 


TRIPLE   INTEGRALS  395 

Example.  Let  a  particle  move  along  a  given  path  in  inter- 
planetary space.  To  find  the  work  done  on  it  by  the  earth, 
supposed  stationary. 

Assume  a  system  of  cartesian  axes  with  the  origin  at  the 
centre  of  the  earth.  Then  the  magnitude  of  the  attraction 
will  be 

\ 


*=v 


and  we  shall  have 


X  =  gcos«  =  ^ 


\     x 


r2    r 

Z  =  gc0Sy=A.?; 
r    r 

J  r8  J  r2        \r0     rj 

Thus  we  see  that  the  work  done  depends  only  on  the 
positions  of  the  extremities  of  O,  not  on  the  particular  path 
joining  the  points,  i.e.  we  have  a  conservative  field  of  force. 

In  connection  with  this  subject  we  will  mention  the  follow- 
ing definition.     Hitherto  we  have  defined  the  definite  integral : 


j 


f{x)  dx 


only  for  the  case  that  a  <  b.  If  a  >  b,  the  definition  is,  how- 
ever, still  valid,  Ax=(b  —  a)/n  now  being  negative.  Hence 
in  all  cases 


Jf(x)dx  =  -Jj 


f(x)  dx. 
Furthermore  we  agree  that 


396  CALCULUS 


a 

ff(x)dx  =  0. 


From  these  relations  we  infer  that 

b  c  b 


Cf(x)dx  =  Jf(x)dx  +  jf(x)dx, 

a  a  c 

no  matter  how  a,  b,  and  c  are  related  to  each  other.     We  can 
also  write : 

b  c  a 

Cf(x)dx+  lf(x)dx  +  ff(x)dx  =  0. 

a  b  c 

EXERCISES 

1.  The  density  of  a  rectangular  parallelopiped  is  propor- 
tional to  the  square  of  the  distance  from  one  vertex.  Find 
its  mass.  Am    Xabc  ^  +  &2  +  ^ > 

o 

2.  Determine  accurately  the  volume  of  the  element  in 
spherical  polar  coordinates,  Fig.  111. 

3.  Find  the  centre  of  gravity  of  the  volume  in  the  preceding 
question. 

4.  Express  the  iterated  integral 

-Va2-x*    2+4X+&U 

dx  I  dy    I  fdz 

0  x+y 

as  a  volume  integral,   and  state  throughout  what  region  of 
space  the  latter  is  to  be  extended. 

5.  The  same  for 


a 


4  2  a  cos  (f> 


I  cos  6  dO  j  sin  <f>d<f>     I    dr. 

_v  0  2  b  cob  <f> 


TRIPLE  INTEGRALS  397 

6.   Write  down  the  five  equivalent  forms  of  the  integral 

;,  y,  z)  dz, 


jdy  jdx  jf(x, 


obtained  by  changing  the  order  of  the  integrations. 

7.  Two  spheres  are  tangent  to  each  other  internally,  and 
also  to  the  x,  y  plane  at  the  origin.  Denoting  the  space 
included  between  the  spheres  by  V,  express  the  volume  integral 


///' 


fdV 


by  means  of  iterated  integrals  in  cartesian  coordinates. 

8.  The  temperature  within  a  spherical  shell  is  inversely 
proportional  to  the  distance  from  the  centre,  and  has  the 
value  T0  on  the  inner  surface.  Given  that  the  quantity  of 
heat  required  to  raise  any  piece  of  the  shell  from  one  uniform 
temperature  to  another  is  proportional  jointly  to  the  volume 
of  the  piece  and  the  rise  in  temperature,  and  that  C  units  of 
heat  are  required  to  raise  the  temperature  of  a  cubic  unit  of 
the  shell  by  one  degree,  find  how  much  heat  the  shell  will  give 
out  in  cooling  to  the  temperature  0°.      Ans.   2irCT0a(b2  —  a2). 

9.  The  interior  of  an  iron  pipe  is  kept  at  100°  C.  and  the 
exterior  at  15°.  The  length  of  the  inner  radius  of  the  pipe  is 
2  cm.,  that  of  the  outer  radius,  3  cm.  The  temperature  at  any 
interior  point  is  given  by  the  formula : 

!T=alogr  +  A 

where  r  is  the  distance  from  the  axis  and  the  constants  a,  fi 
are  to  be  determined  from  the  above  data.  Taking  the  specific 
heat  of  iron  as  .11,  and  its  specific  gravity  as  7.8,  how  much 
heat  will  a  segment  of  the  pipe  30  cm.  long  give  out  in  cooling 
to  0°  ?  Ans.   21,000  calories. 

10.  Determine  the  attraction  of  a  bar,  of  rectangular  cross- 
section,  on  an  exterior  particle  situated  in  its  axis. 


CHAPTER  XX 
APPROXIMATE  COMPUTATIONS.    HYPERBOLIC  FUNCTIONS 

1.  The  Problem  of  Numerical  Computation.  It  frequently 
happens  in  practice  that  we  wish  to  know  the  value  of  a  func- 
tion for  a  special  value  of  the  independent  variable  or  that  we 
wish  to  compute  a  definite  integral.  In  all  such  cases  only  a 
limited  number  of  decimal  places  or  of  significant  figures, 
as  the  case  may  be,  are  of  interest  in  the  result,  for  the  data 
of  the  problem  are  accompanied  by  errors  of  observation  or  are 
otherwise  inexact,  and  as  soon  as  these  errors  begin  to  make 
themselves  felt,  we  have  obviously  reached  the  limit  of  accu- 
racy for  the  result  in  hand.  Hence  any  method  that  will 
enable  us  to  obtain  the  result  with  the  degree  of  accuracy 
above  indicated  yields  a  solution  of  our  problem. 

On  the  other  hand,  rough  approximate  solutions  of  the  kind 
we  are  about  to  take  up  serve  as  useful  checks  for  solutions 
obtained  by  other  methods. 

2.  Solution  of  Equations.    Known  Graphs. 

Example*   Let  it  be  required  to  solve  the  equation 

(1)  cos  x  +  \x  =  0,  0<X<ir. 
The  student  has  constructed  the  graph  of  the  curve 

(2)  y  ==  cos  x 

*  This  example  and  the  exercise  present  themselves  in  the  following 
problem  of  Mechanics.    A  heavy  uniform  circular  disc  can  turn  freely 

398 


APPROXIMATE   COMPUTATIONS  399 

accurately  to  scale.     Since  equation  (1)  is  equivalent  to  the 

equation 

(3)  —  Ja;  =  cosic, 

we  can  obviously  formulate  our  problem  as  follows.     To  find 
the  intersection  of  the  curves : 

y—  —  \x,  y  =  cosx. 

Graphically,  then,  it  will  be  sufficient  to  draw  a  straight  line 
through  the  origin  and  the  point  x  =  4,  y  ss  —  1,  and  observe 
the  abscissa  of  the  point  of  intersection  with  the  cosine  curve. 


EXERCISE 

Find  the  largest  value  of  P  for  which  the  equation : 
cosx  +  Px  =  0 
admits  a  solution  in  the  interval  0<  x<ir. 

3.  Newton's  Method.   Let  it  be  required  to  solve  the  equation 

(4)  /(z)  =  0. 

In  practice  we  usually  know  that  the  equation  has  a  solution 
within  a  restricted  interval.  Moreover,  f(x)  will  be  a  continu- 
ous function  in  this  interval,  and  its  derivative  will  not  vanish 
there.  We  can  frequently  make  a  good  guess  at  the  solution 
to  begin  with.  Take  this  value,  x  =  al9  as  a  first  approxima- 
tion. Then  we  shall  get  a  second  approximation  if  we  draw 
the  tangent  at  the  point  x  =al}  y  =f(a1)  and  take  the  point 
x  =  a2  in  which  the  tangent  cuts  the  axis  of  x,  y  —  0,  Fig.  117. 
The  equation  of  the  tangent  in  question  is 

about  a  horizontal  axis  through  its  centre,  perpendicular  to  its  plane. 
There  is  a  weight  W  fastened  to  the  rim  of  the  disc  and  a  fine  thread  is 
wound  round  the  rim  and  hangs  down,  carrying  a  weight  Q  at  its  end. 
W  being  at  the  lowest  point  of  the  disc  and  the  free  end  of  the  string 
being  vertical,  the  system  is  released.  Find  how  high  W  will  rise  and 
determine  the  least  value  of  IF  for  which  W  will  not  be  pulled  over. 


400  CALCULUS 

y  _/(a1)=/'(a1)  (x-a,). 

For  its  point  of  intersection  with,  the  axis 
of  x: 

^2     «i  0  -/(ax)  =/  («!)(»-%). 

Fig.  117  f(    . 

Hence  a,  =  ^-4^. 

To  get  a  third  approximation,  proceed  with  a2  as  above  with 
ax,  and  so  on. 

If  fix)  is  a  polynomial  with  numerical  coefficients,  the  actual 
computation  of  /(«i)  and  /'  (ax)  would  be  laborious.  To  meet 
this  difficulty  Horner's  Method  has  been  devised,  cf.  any  of  the 
standard  text-books  on  Higher  Algebra. 

Example.  It  is  shown  that  the  equation  of  the  curve  in 
which  a  chain  hangs,  —  the  Catenary,  —  is 


(5)  y  =  \\f+e 

where  a  is  a  constant.     The  length  of  the  ai  c,  measured  from 
the  vertex,  is 

(6)  s<{°"-e~l) 

Let  it  be  required  to  compute  the  dip  in  a  chain  32  feet  long, 
its  ends  being  supported  at  the  same  level,  30  feet  apart. 

We  can  determine  the  dip  from  (5)  if  we  know  a,  and  we 
can  get  the  value  of  a  from  (6)  by  setting  s  =  16,  x  =  15 : 


16 


1  ^ 
Letz=  —  •     Then 


15  15 


f(x)  =  ex-  e~x  -  ffa  =  0, 
and  we  wish  to  know  where  the  curve 


APPROXIMATE   COMPUTATIONS  401 

(7)  y=f(x)  =  e*-e-*-iix 

crosses  the  axis  of  x. 

This  curve  starts  from  the  origin  and,  since 

^  =  f'(x)=:e*  +  e-*-i% 

is  negative  for  small  values  of  x,  the  curve  enters  the  fourth 
quadrant.     Moreover, 

g  =  e*-e-*>0,  x>0, 

and  hence  the  graph  is  always  concave  upward.     Finally, 

/(l)  =  e-e-'-2T%  =  .2ir>0, 

and  so  the  equation  has  one  and  only  one  positive  root  and 
this  root  lies  between  0  and  1. 

It  will  probably  be  better  to  locate  the  root  with  somewhat 
greater  accuracy  before  beginning  to  apply  the  above  method. 
Let  us  compute,  therefore,  /(J).  By  the  aid  of  the  Tables, 
p.  121,  we  find : 

/(.5)  =  1.6487  -  .6065  - 1.0667  =  -  .0245  <0. 

Comparing  these  two  values  of  the  function  : 

/(.5)=-.02,  /(1)=.22, 

and  remembering  that  the  curve  is  concave  upward,  so  that 
the  root  is  somewhat  larger  than  the  value  obtained  by  direct 
interpolation  (this  value  corresponding  to  the  intersection  of 
the  chord  with  the  axis  of  x)  we  are  led  to  choose  as  our  first 
approximation  ax  =  .6 : 

/(.6)  =  1.8221  -  .5488  - 1.2800  =  -  .0067, 

/(.6)  =r  1.8221  +  .5488  -  2.1333  =      .2376, 

a2  =  .6-  ~  -0067  =  .6  +  .0282  =  .628. 
2  .2376 

To  get  the  next  approximation,  a3,  we  compute 

/(.628)  =  1.8739  -  .5337  - 1.3397  =  .0005. 
2d 


402 


CALCULUS 


Hence  the  value  of  the  root  to  three  significant  figures  is  .628 
with  a  possible  error  of  a  unit  or  two  in  the  last  place,  and  the 
value  of  a  we  set  out  to  compute  is,  therefore,  15/.628  =  23.9. 

4.  Direct  Use  of  the  Tables.  While  explaining  methods  of 
solution  more  or  less  obvious  geometrically,  we  must  not  over- 
look an  immediate  solution  of  the  problem  in  certain  cases  by 
mere  inspection  of  the  tables. 

For  example,  the  equation 

cos  x  =  x 

has  one  and  only  one  root,  as  we  see  by  inspection  of  the 
graphs  of 

y  =  cos  x        and        y  =  x. 

To  find  this  root,  turn  to  the  Tables,  p.  134.     There  we  find : 


RADIANS 

DEGREES 

COSINES 

.7418 

42°  20' 
42°  30' 

.7392 19 
.7373 

Hence  x  =  .7391,  corresponding  to  an  angle  of  42°  21'.  The 
interpolation  by  which  x  was  found  is  a  neat  problem  in  ele- 
mentary algebra.     It  is  left  to  the  student. 

The  example  of  §  3  is  nearly  a  case  in  point.     The  hyper- 
bolic sine  and  cosine  are  defined  by  the  equations  (cf.  §  8) : 


sh#  = 


ex  — e~ 


ch#  = 


ex  +  e~x 


Tables  of  values  of  these  functions  are  given  on  pp.  120-123 
of  the  Tables.  The  problem  in  hand  thus  becomes  the  fol- 
lowing :  to  solve  the  equation 

shsc  =  -j-f  cc. 

From  the  tables  on  p.  121  we  find : 


APPROXIMATE   COMPUTATIONS  403 


X 

H* 

sha; 

.62 
.63 

.6613 
.6720 107 

.6605 

n„n„  120 

.6725 

Hence  the  value  of  x  given  by  direct  interpolation  is  .626. 


Solve  the  equation : 


EXERCISE 


cot  X  =  X. 


5.  Successive  Approximations.  The  method  of  successive 
approximations  is  most  easily  understood  by  inspection  of  the 
graphs  of  the  functions.  There  are  two  cases,  both  illustrated 
in  the  accompanying  figures.  If  the  slopes  of  the  curves  both 
have  the  same  sign,  let 


Q:  F(x,y)=0 

be  the  one  that  is  less  steep, 
C2: 


or 


3>  (x,  y)  =  0        or 


V  =f(x) 

x  =  <f>(y) 


Cd 

/Ci 

11 

Jf\ 

^ 
\ 

A    j 

' 

X 

0 

'• 

Cj  x2 

y 

<?>\ 

r"^1 

43 

X 

0 

x2  .; 

cl 

Fig.  118 

the  other.     Then,  making  the  best  guess  we  can  to  start  with, 
x  =  x1,  compute 

2/i  =/Oi) 
and  substitute  this  value  in  the  equation  of  C2,  thus  getting 
the  second  approximation: 

a2  =  <Kvi)- 

Proceeding  with  x2  in  the  same  manner  we  obtain  first  y2 , 
then  x3,  and  so  on. 


404 


CALCULUS 


The  successive  steps  of  the  process  are  shown  geometrically 
by  the  broken  lines  of  the  figures. 

The  success  of  the  method  depends  on  the  ease  with  which 
y  can  be  determined  when  x  is  given  in  the  case  of  Glt  while 
for  C2  x  must  be  easily  attainable  from  y.  If  the  curves  hap- 
pened to  have  slopes  numerically  equal  but  opposite  in  sign, 
the  process  would  converge  slowly  or  not  at  all. 

The  method  has  the  advantage  that  each  computation  is 
independent  of  its  predecessor.  An  error,  therefore,  while  it 
may  delay  the  computation,  will  not  vitiate  the  result. 

Example.  A  beam  1  ft.  thick  is  to  be  inserted  in  a  panel 
10  x  15  ft.  as  shown  in  the  figure.  How  long  must  the  beam 
be  made  ? 

We  have : 

sin  <j>  + 1  cos  <f>  =  15, 

cos  <£  -f- 1  sin  <f>  =  10. 

Hence     cos2<£  —  sin2<£=10  cos  <£  — 15  sin  <f>. 

Now  an  expression  of  the  form 

a  cos  cji  —  b  sin  <j> 


Fig.  119 


can  always  be  written  as 

VaT+tf( cos  <j>  - 

Wa2  +  &2 

where  cos  a  — — 


Va2+62 


sin  <£  )  =  Va2 + b'2  cos  (<£  -h  a), 


sma  = 


Va2  +  &2  Va2  +  62 

In  the  present  case,  then : 

cos  2  <f>  =  V325  cos  (<£  -f  a), 


where 


cos  a  — 


10 


sma  = 


15 


V325  V325 

Thus  a  is  an  angle  of  the  first  quadrant  and 
tan  a  =  4,  a  =  56°  16'. 


APPROXIMATE  COMPUTATIONS  405 

Our  problem  may  be  formulated,  then,  as  follows :   To  find 
the  abscissa  of  the  point  of  intersection  of  the  curves : 

y  —  cos  2  <f>,  y  =  V325  cos  (<j>  4-  a). 

We  know  a  good  approximation  to  start  with,  namely : 

tan  <£  =  £,  <£  =  33°44'. 

For  this  value  of  <f>  the  slopes  are  given  by  the  equations : 

^.^  =  -2sin2d>=-2sin67°28f=-1.8, 
■k     d<l>  ' 

1??.-^=-  V325sin(d,-f  a)  =-  V325  =  -18. 
ir     d<j> 

Hence  we  have : 

Cx :  y  —  cos  2  <f> ; 

(L :   y  =  V325  cos  (d>  +  a)         or         <f>  =  cos  *  — %==  —  a. 

V325 
Beginning  with  the  approximation 

<fo  =  33°44', 

we  compute  y1  —  cos  67°  28'. 

Passing  now  to  the  curve    C2,   we  compute  its   <f>   when   its 

2/  =  2/i: 

yx  =  V325  cos  (<£2  +  a),  <f>2  =  32°  31'. 

We  now  repeat  the  process,  beginning  with  <£2  =  32°  31'  and 

find: 

2/2  =  cos  65°  02', 

y2  =  V325  cos  (fc  +  a),  <f>3  =  32°  23'. 

A  further  repetition  gives  <£4  =  32°  22',  and  this  is  the  value 
of  the  root  we  set  out  to  determine. 


406  CALCULUS 

EXERCISES 

1.  Solve  the  same  problem  for  a  beam  2  ft.  thick. 

2.  A  cord  1  ft.  long  has  one  end  fastened  at  a  point  0  2  ft. 
above  a  rough  table,  and  the  other  end  is  tied  to  a  rod  2  ft. 
long.  How  far  can  the  rod  be  displaced  from  the  vertical 
through  0  and  still  remain  in  equilibrium  when  released  ? 

The  equations  on  which  the  solution  de- 
pends are : 

2cot0  +  -  =  cot<fc 


P 


FlG'120  [       2cos0  +  cos4>  =  2. 

If  the  coefficient  of  friction  /x  =  ^,  find  the  value  of  cf>. 

3.  A  heavy  ring  can  slide  on  a  smooth  vertical  rod.  To 
the  ring  is  fastened  a  weightless  cord  of  length  2  a,  carrying  an 
equal  ring  knotted  at  its  middle  point  and  having  its  further 
end  made  fast  at  a  distance  a  from  the  rod.  Find  the  position 
of  equilibrium  of  the  system. 

4.  Solve  the  example  worked  out  in  §  3  by  the  method  of 
successive  approximations. 

5.  In  the  example  worked  in  the  text  replace  cos  <j>  by  its 
value  in  terms  of  sin<£,  reduce  the  resulting  equation  to  the 
form  of  an  algebraic  equation  in  sin  <j>  and  solve  the  latter  by 
Horner's  Method. 

6.  Definite  Integrals.  Simpson's  Rule.  If  we  wish  actually 
to  compute  the  area  under  a  curve  numerically,  we  can  make 
an  obvious  improvement  on  the  method  of  inscribed  rectangles 
by  using  trapezoids,  as  shown  in  Fig.  53.  We  begin  as  before 
by  dividing  the  interval  (a,  b)  into  n  equal  parts,  and  we  denote 
the  length  of  each  part  by  h.     The  area  of  the  k-th  trapezoid  is 

i(y*+-y»+i)* 

and  hence  the  approximation  thus  obtained  is 


APPROXIMATE   COMPUTATIONS  407 

This  formula  is  known  as  the  Trapezoidal  Rule.  If  the  curve 
is  concave  downward,  as  in  Fig.  53,  Ax  is  too  small. 

Again,  if  we  take  n  as  an  even  integer  and  draw  tangents  at 
the  points  (xu  yx),  (x3,  ys),  •••  (#„_!  y«_i),  we  get  some  trapezoids 
as  shown  in  the  figure,  the  area  of  any  one  being  2ykh,  where 
k  is  odd.     Hence  ^r^r 

A2  =  2h[yl  +  y3-\ hy—i] 

'vlx  i  Vm 
is  an  approximation  which  is  too  large,  and 

AX<A<A2.  ^  FlQ  121 

If  the  curve  is  concave  upward,  the  inequalities  must  be 
reversed. 

Finally,  a  still  closer  approximation  may  be  obtained  by 
using  arcs  of  parabolas  instead  of  straight  lines.  If  we  make 
the  parabola 

y  =  a  +  b  (x  —  xk)  +  c(x  —  xk)2 

go  through  three  successive  points,  (%._0  yk-i),  (xkt  yk), 
(xk+l,  yk+i),  it  will  follow  the  arc  of  the  curve  more  closely  in 
between  than  the  broken  lines  or  the  tangents  of  the  preceding 
approximations  do.     Now  the  area  under  the  parabolic  arc  is 

xk+h 


I' 


[a  +  b  (x  —  xk)  +  c  (x  —  xk)2]  dx= 


and  it  remains  to  determine  a  and  c  from  the  above  conditions : 

x  =  xk  j  IJk  —  a  '•> 

x  =  xk  +  h,  yk+1  =  a  +  bh  +  ch?', 

x  =  xk  —  h,  yk_x  =  a  —  bh  +  ch?. 

Hence              a  =  yk ,  2ch2  =  yk_1-2  yk  -f  yk+1 . 

Thus  the  area  under  the  parabolic  arc  is  seen  to  have  the 
value 


408  CALCULUS 

iftfot-i  +  ^t  +  ft+i). 
Adding  these  areas  for  k  =  l9  3,  •••  n  —  1,  we  get  a  new  ap- 
proximation : 

^3  =  P[2/o  +  2/n  +  2G/2  +  2/4+  •••2/„-2)  +  4(2/1  +  2/3+  —  +|f«-i)]. 
This  formula  is  known  as  Simpson's  Rule. 
If  we  set  u  =  y0  +  yn, 

v  =  yi  +  Vs+  —  4-2/n-i,  ™  =  2/2  +  2/H by— s, 

we  have:      A1  =  ^h(u  +  2v  +  2w),         A2  —  2hvf 

A3  =  ±h(u+4:V  +  2w). 
It  turns  out  that  A3  =  %A1  +  ±A2. 

2 

Example.*    Consider   / — ,  and  let  n  =  10.    Then  h  =  .  land 

it  =  1.5,        v  =  3.459  539  4,         to  =  2.728  174  6. 

Hence  ^  =  .693  771,         A2  =  .691  908,        A3  =  .693  150. 

The  value  of  the  integral  is  ( Tables,  p.  109) : 

log  2  =  .693  147. 

Thus  Ax  differs  from  the  true  value  by  less  than  7  parts  in 
about  7000,  or  one  tenth  of  one  percent.  A2  differs  by  about 
12  parts  in  7000 ;  while  A3  is  in  error  by  less  than  3  parts  in 
600,000,  or  1  part  in  200,000. 

l  EXERCISES 

1.   Compute    I  exdx;  taking  w  =  10,  and  compare  the  result 


0 

with  that  obtained  by  integration.    Note  the  tables  on  pp.  120, 
121  of  the  Tables. 

*  These  figures  are  taken  from  Gibson's  Elementary  Treatise  on  the 
Calculus,  p.  331,  to  which  the  student  is  referred  for  further  examples. 
A  more  extended  treatment  of  the  subject  of  this  paragraph  will  be  found 
in  Goursat-Hedrick,  Mathematical  Analysis,  vol.  1,  §  100. 


APPROXIMATE  COMPUTATIONS  409 

/dx      ' 

1 

3.  Obtain  an  approximate  formula  for  the  content  of  a  cask 
whose  bung  diameter  is  a,  head  diameter,  b,  and  length,  I. 

Ans.    ^[8a2  +  4a&+.3&2]. 

4.  If  in  the  preceding  question  a  is  only  slightly  greater 
than  b,  the  formula  may  be  replaced  by  the  simpler  one : 

iral(a  +  2b). 

7.  Amsler's  Planimeter.  A  curve  may  be  given  graphically, 
as  in  naval  architecture,  when  the  plans  of  a  ship  are  made  by 
drawing  to  scale  successive  cross-sections.  Again,  take  the 
indicator  diagrams  of  a  steam  engine.  A  pencil  or  stylus  is 
carried  over  a  sheet  of  paper,  tracing  a  curve  as  shown  in 
Fig.  122.  The  height  of  the  pencil  above  the  axis  of  abscissas 
represents  the  pressure  p  of  the  steam  on  the  piston,  and  the 
abscissa  is  proportional  to  the  distance  the 
piston  has  travelled.  Hence  the  work  done  p  (  \* 
in  the  direct  stroke  is  proportional  to  * 

pdx, 


f> 


the  ordinate  p  being  given  by  the  upper  part  of  the  curve. 
When  the  piston  returns,  negative  work  is  done,  and  the 
amount  is 


-  I pdx        or  I pdx, 


Since  x  is  proportional  to  the  volume  of  steam  behind  the  piston,  we 
may  also  write  the  work  as 

pdv. 


J 


410  CALCULUS 

the  ordinate  now  being  given  by  the  lower  part  of  the  curve. 
Hence  the  total  work  done  is  proportional  to  the  algebraic  sum 
of  these  two  integrals,  namely,  the  line  integral 


I  pdx        or  I  pdv, 


taken  round  the  complete  boundary,  i.e.  the  work  is  propor- 
tional to  the  area  enclosed  by  the  curve. 

In  order  to  compute  such  areas  one  method  is  that  of  §  6, 
and  this  is  the  one  employed  in  naval  architecture.  Another 
method  is  by  means  of  integrating  machines,  integraphs,  or 
planimeters,  as  they  are  called,  and  this  is  the  one  employed 
for  measuring  indicator  diagrams.  There  are  several  such 
machines  in  use,  one  of  which,  Amsler's  Planimeter,  we  will 
now  describe.  It  consists  of  two  arms,  OP  and  PQ,  jointed  at 
P.  One  arm  is  pivoted  at  0 ;  the  other  has  a  point  at  its  end 
Q,  and  Q  is  made  to  trace  out  the  curve  whose  area  is  sought. 

The  theory  is  as  follows.     Consider  the  area  swept  out  by 

the  arm  PQ>     Give  to  this  arm  an  infinitesimal  displacement, 

its  new  position  being  P'Q'.     The  corresponding  infinitesimal 

z-^   ,         increment  of  area,  AA,  is  seen  to 

(  yts       differ   from   the   area  PQSQ'P'P, 

sC^s®      where  SQ  is  congruent  to  the  arc 

y^^h\C    J       PP'  &&&  makes  the  same  angle  with 

jjj*  ays  PQ,  by  an  infinitesimal  of  higher 

•/^\/($L order.     But  this  latter  area  is  ob- 

^^cP viously  equal  to 

FlGl23  Ih +  ±  I2  Acf>, 

where  h  denotes  the  perpendicular  distance  from  P'S  to  PQ 
and  I  is  the  length  of  PQ.     Hence 

AA  =  lh  +  %l2A<f>  +  c, 

where  c  is  an  infinitesimal  of  higher  order. 

In  order  to  measure  h,  a  disc  is  attached  to  the  arm  PQ  at  R, 
the  axis  of  the  disc  coinciding  with  that  arm.*     The  disc  can 

*  As  a  matter  of  fact,  B  lies  in  the  line  QP  produced.  This  alters 
nothing  in  the  theory,  the  distance  PR  =  a  merely  being  taken  negative. 


APPROXIMATE   COMPUTATIONS  411 

turn  freely  on  its  axis  and  the  rim  of  the  disc  rests  on  the 
paper.  Now  suppose  that  the  arm  PQ  were  brought  into  its 
new  position  P'Q'  as  follows: 

(a)  PQ  is  moved  in  its  own  line  till  P  reaches  the  foot  of 
the  perpendicular  dropped  from  P  on  its  line ; 

(b)  PQ  is  moved  perpendicular  to  itself  till  it  comes  into  the 
position  PS ; 

(c)  PQ  is  rotated  about  P'  as  a  pivot  till  it  comes  into  the 
final  position  P'Q'. 

It  is  now  easy  to  compute  the  angle  through  which  the  disc 
has  turned.  During  the  movement  (a)  it  does  not  turn  at  all. 
During  (b)  it  turns  through  an  angle  proportional  to  h,  h/r, 
where  r  is  the  radius  of  the  disc ;  and  during  (c)  through  an 
angle  aA<f>/r,  where  a  denotes  the  length  PP.  The  total  angle 
thus  obtained,  (/i  +  aA</>)/r,  will  differ  from  the  angle  Aw  due 
to  the  actual  displacement  at  most  by  an  infinitesimal  of  higher 
order,  rj :  « 

A         h-\-aAd>  . 

Aa>  sx  — £ £  -f-  ri. 

r 

This  assumption  is  an  axiom  or  physical  law,  borne  out  by 
experience,  on  which  the  whole  theory  of  this  machine  rests. 

If  we  eliminate  h  between  the  equation  for  AA  and  that  for 
Aw,  we  get : 

A  A  =  ZrAw  +  (1Z2  -  al)  A<£  -  lrv  +  c. 

Dividing  by  A<£  and  allowing  A<£  to  approach  0  as  its  limit,  we 
obtain : 

D^A^lrD^+^-al) 

and  hence  dA  =  Ir d<o  +  (i  I2  —  al)  d<f>. 

The  simplest  case  is  that  in  which,  as  Q  describes  the  closed 
curve  in  question,  <£  steadily  increases  for  one  arc  from  <£0  to 
<£i  and  steadily  decreases  for  the  remaining  arc  from  fa  to  <£0. 
The  total  area  swept  out  for  the  first  arc  is 

Ax  =  ^-(wj  —  w0)  -h  (-J-Z2  —  al)  (fa  —  <t>0). 


412  CALCULUS 

For  the  second  arc,  <£  is  decreasing,  and  the  area  will  be 
negative : 

A2  =  lr(Q  -  cuO  +  (il2  -  oZ)(^o- ^)- 

The  area  of  the  curve  is  the  algebraic  sum  of  these  two 

areas : 

A  =  AX  +  A2  =  Ir  (O  —  o>o) 

and  hence  is  proportional  to  the  angle  O  —  o>0  through  which 
the  disc  has  turned.  This  angle  is  read  off  on  the  vernier,  and 
the  constant  multiplier  is  known  or  determined  for  the  par- 
ticular machine  that  is  being  used. 

It  can  be  shown  generally  that  the  area  of  any  closed  curve 
is  given  by  the  same  formula,  provided  <f>  comes  back  to  its 
initial  value,  the  method  being  merely  to  divide  the  area 
enclosed  by  the  curve  up  into  pieces,  for  each  of  which  the 
above  determination  is  applicable.  But  if  the  bar  PQ  makes 
a  complete  rotation,  so  that  cj>  changes  by  2-rr,  the  integral  of 
the  last  term  in  the  expression  for  dA  will  not  vanish,  but  will 
contribute  (-J- I2  —  at)  •  2  it  to  the  result. 

8.  The  Hyperbolic  Functions.  Certain  functions  analogous 
to  the  trigonometric  functions,  called  the  hyperbolic  functions, 
have  recently  come  into  general  use.  They  go  back,  however, 
to  Riccati  (1757)  and  are  defined  as  follows : 

.   ,  ex-e~x 

sinn  x  = 


2 

e*  +  e-* 
cosh  x  =  — l- — 

m 

sinha; 


tanh  x 


coshx' 


etc.  (read  "  hyperbolic  sine  of  x,"  etc.).  An  abbreviated  nota- 
tion for  sinh  x,  cosh  x,  tanh  x,  is  sh  x,  ch  #,  th  x.  The  graphs  of 
these  functions  are  shown  in  Fig.  124.  The  functions  satisfy 
the  following  functional  relations,  sh#  and  tho;  being  odd 
functions,  ch  x  an  even  function : 


HYPERBOLIC   FUNCTIONS 


413 


sh(—  x)=  —  sh^,     ch(—  x)  =  ch#,     th(— £)=  — th#. 
Moreover:  shO  =  0,         chO  =  l,         thO  =  0. 

Also :  ch2  x  —  sh2  x  —  1, 

1  —  th2  x  =  sech2#,  coth2  a;  —  1  =  csch2a;. 

?/  =  th.r 

ttL 


Fig.  124 

The  Addition  Theorems  are  as  follows : 

sh  (cc  +  ?/)  =  sh  ic  ch  ?/  -f-  ch  ^  sh  y ; 
ch  (x  +  y)  =  ch  x  ch  ?/  +  sh  x  sh  # ; 

1 1  /     ,     x       th  x  4-  th  y 

th  (x  +  w)  = — f—  • 

/    V         ;      1  +  thajthy 

From  these  relations  follow  at  once : 

sh2#  =  2sh£chx, 

ch  2x  =  ch2x  +  sh2a  =  2  ch2a  -1  =  1  +  2sh2a,\ 

Derivatives  of  the  Hyperbolic  Functions.      The   derivatives 

have  the  values: 

d  sh  x       ,  d  ch  x       , 

=  ch  #,     =  sh  x, 


dx 


dthx 
dx 


=  sech2aj, 


dx 
d  coth  x 


dx 


=  —  csch2#, 


etc. 


Tlie  Inverse  Functions.     The  inverse  of  the  hyperbolic  sine 
is  called  the  antihyperbolic  sine : 

y  =  ah~1x         if         x  =  shy. 
Hence  ««-.}(*  —  *-*). 


414  CALCULUS 

Solving  for  ep,  we  get : 


ev  =  x  ±  vT+a?. 
Since  ey  >  0  for  all  values  of  y,  the  upper  sign  alone  is  possible, 


and  y=sh-1x  =  \og(x-\-^/l-{-x2). 

The  antihyperbolic  cosine,  however,  is  multiple-valued,  as 
appears  from  a  glance  at  its  graph,  obtained  as  usual  in  the 
case  of  an  inverse  function  by  rotating  the  graph  of  the  direct 
function  about  the  bisector  of  the  angle  made  by  the  positive 
coordinate  axes : 


ch_1a;  =  log  (a;  ±  s/x2  —  1),  x^l. 

The  upper  sign  corresponds  to  positive  values  of  ch"1**;. 

Also:  th"1a;  =  -ilogi±^,         -l<a;<l. 

1  —  x 

The  derivatives  have  the  values: 
d  sh_1a;  1 


dx          Vl  +  x2' 
d  ch-1a; 


dx  Va^-l' 

dth-1^     1 
dx         1—x2 

We  thus  obtain  a  close  analogy  between  certain  formulas  of 
integration : 

/dx  .     ,a>  C     dx  ,  _,x 

—  = :sin  1-.  f  —  =sh  l-; 

Vtf-x2  a         JVa2  +  «2  a 

/*  dx        1 ,     _,x  C   dx        1  .v_i» 

— — ^-tan"1-,  I- — 3  =  -th  »-. 

cr  +  ar     a  a         J  a2  — or     a        a 

A  collection  of  formulas  relating  to  the  hyperbolic  functions 
will  be  found  in  Peirce's  Tables,  pp.  81-83,  and  tables  for 
shx  and  ch#  are  given  there  on  pp.  119-123. 


HYPERBOLIC   FUNCTIONS  415 

Relation  to  the  Equilateral  Hyperbola.     The  formula: 


Vl  —  3?dx  ==  \  x  Vl  —  x2  + 1  sin"1 


expresses  the  area  OQPA  under  a  circle  in  terms  of  the  function 
sin-1#  and  enables  us,  on  subtracting  the  area  of  the  triangle 
OQP  from  each  side  of  the  equation,  to  interpret  sin-1  a;  as 
twice  the  area  of  the  circular  sector  OP  A. 

y 

y      \    \     x 
Q 

Fig.  125 

There  is  a  similar  interpretation  for  sh_1a;  with  reference  to 
the  equilateral  hyperbola 

y2  =  l  +  x?. 


J 


Vl  +  x2 dx  =  %x VT+aT2  +  i  log (x  +  VT+xV 


i^Vl+^  +  Jsh-1^. 


Thus  we  see  that  sh_1#  is  represented  by  twice  the  area  of  the 
hyperbolic  sector,  OPA. 

To  the  formulas  for  the  circle : 

&  +  y2  =  l, 

x  =  sin  u,  y  =  cos  u, 

correspond  the  following  formulas  for  the  hyperbola: 

x  —  sh  u,  y  =  ch  u, 

the  parameter  u  being  represented  geometrically  in  each  case 
by  twice  the  area  of  one  of  the  above  sectors. 


416  CALCULUS 

The  analogy  of  the  hyperbolic  functions  to  the  trigonometric 
functions  is  but  one  phase  of  the  fact  that  in  the  domain 
of  complex  quantities  the  trigonometric  and  the  exponential 
functions  and  their  inverse  functions,  the  ant itrigonome trie 
functions  and  the  logarithms  are  closely  related.  We  have 
already  had  occasion  to  mention  the  formula: 

e*'  =  cos  <f>  +  i  sin  <f>,  i  =  V  — •  1. 

Thus  sin  z  = 


2i 

zi    i_  a—zi 


cos  z  = — 


sin-1  z  =  -  log  (zi  ±  Vl  —  z2), 
i 

tan  i*  =  _log— , 

where  2!  =  #  +  y£  is  any  complex  quantity . 

2%e  Gudermannian.     Let  <£  be  defined  as  a  function  of  x  by 
the  relation : 

sh#  =  tan<£,         <£  =  tan-1sh#,         —-<<t><Z- 

Then  <f>  is  called  the  Gudermannian  of  #  and  is  denoted  as 
follows :  * 

<£=gdx. 


We  have : 

sh#  = 

=  tan  <f>, 

chx  = 

:  sec  <£, 

thx 

=  sin 

<t>, 

csch  x  = 

=  COt  <j>, 

sech  x  = 

:  COS  <£, 

cotha? 

=  csc 

4>; 

and  since 

ex  =  ch 

a;  +  sh  a;, 

ex-. 

=  tanfc 

+i) 

a;  =  log  tan  f  -  + 

V4 

!)■ 

*  Also  called  the  hyperbolic  amplitude  and  denoted  by 

amhse. 

APPENDIX 

A.  — THE  EXPONENTIAL  FUNCTION 

In  Chap.  II,  §  8,  it  was  shown  that,  when  x  =  a  >  1, 

(1)  an'>an         if         ri>n, 

where  n  and  n'  are  two  positive  or  negative  rational  numbers. 
Moreover 

(2)  a">0 

for  all  rational  values  of  n ;  and 

(3)  lim  aM  =  +  oo  ,  lim  an  =  0. 

n=+»  n=— oo 

One  further  relation,  which  we  will  now  prove,  is  important, 
namely: 

(4)  lim  an  ==  1. 

M  =  0 

When  0<  n  <  1,  the  curve 

y  =  xn 

is  concave  downward,  for  Dx2y  =  n(n  —  l)af~2<0,  and  so  it 
lies  below  its  tangent.  The  equation  of  the  latter  in  the 
point  (1,  1)  is : 

y  =  n(x-l)+l. 

Hence  for  such  values  of  n,  the  ordinate  of  the  curve,  an,  is 
less  than  the  ordinate  of  the  tangent,  n  (a  —  1)  -f  1 : 

1  <  an  <  n  (a  - 1)  + 1,  0  <  n<  1. 

2e  417 


418  CALCULUS 

Thus  (4)  is  seen,  at  least,  to  be  true  whe«  n  approaches  0 
from  the  positive  side. 

Similarly  it  is  shown  that,  when  n<0,  the  curve  is  concave 
upward,  and 

l>an>n(a  — 1)4-1,  n<0. 

Hence  (4)  is  true  when  n  approaches  0  from  the  negative  side, 
too,  and  the  relation  is  thus  established  generally. 
We  can  now  prove  the  following  theorem. 

Theorem  1.  Ifvbe  any  irrational  number  and  n  be  allowed  to 
approach  v,  passing  only  through  rational  values,  then  an  approaches 
a  limit. 

First,  let  n  approach  v  from  below,  n<v.  Then,  by  (1), 
an  steadily  increases  as  n  increases,  but  never  becomes  so  great 
as  a1',  where  V  is  any  rational  number  greater  than  v.  Hence, 
by  the  Fundamental  Principle  for  the  existence  of  a  limit, 
Chap.  XII,  §  3,  an  approaches  a  limit  not  greater  than  a1',  and 
in  fact  here  less.  For,  if  I"  be  chosen  between  v  and  V,  then 
lim  an  is  not  greater  than  a1",  and  a1"  <a1'.  Denote  the  limit 
by  A.     Then 

(5)  lim  an  =  A  <  a1',  V  >  v. 

n=v— 

Here,  V  is  any  rational  number  greater  than  v. 

Again,  let  n  approach  v  from  above,  n  =  n'  >  v.  Then,  by 
similar  reasoning,  anf  approaches  a  limit  A1  >  a1,  where  I  is  any 
rational  number  less  than  v. 

(6)  lim  an'  =  A'  >  a1,  l<v. 

n'=v+ 

Finally,  to  show  that  A'  =  A.  It  is  clear  that  A'  is  not 
less  than  A,  for  (5)  gives,  when  V  =  oo  : 

A<A'. 

Since  a1'  >  A'  and  a1  <  A,  we  infer  that 

av-al>A'-A. 

Setting  V  =  1 4-  h,  we  get : 

Q^A'-A<a\ah-1). 


THE  EXPONENTIAL   FUNCTION  419 

Now  let  I  and  V  both  approach  v.  Then  h  approaches  0  and 
the  right-hand  member,  therefore,  approaches  0.  But  A  and 
A'  do  not  change  with  I  and  V,  and  so  the  value  of  their  differ- 
ence, being  constant,  must  be  0 : 

0=A'-A. 

This  completes  the  proof. 

Definition.  For  an  irrational  value  of  the  exponent,  n  =  v, 
we  will  define  av  as 

lim  an, 

n  passing  through  only  rational  values. 

Relations  (l)-(4)  are  readily  shown  to  hold  when  n  and  n' 
are  one  or  both  irrational. 

Theorem  2.     The  function 

thus  defined  is  continuous. 

We  wish  to  prove  that,  if  xQ  is  an  arbitrary  value  of  x,  then 

lim  of  sa  ax°. 

x±x0 

The  proof  is  similar  to  that  of  Theorem  1 ;  but  the  present 
theorem  differs  from  that  one  in  that  x0  is  any  number,  rational 
or  irrational,  and  furthermore  x,  in  approaching  x0,  passes 
through  all  values,  irrational  as  well  as  rational. 

First  let  x  approach  x0  from  below,  x<  x0.  Then  it  follows 
as  in  the  proof  of  Theorem  1  that  ax  approaches  a  limit  A : 

lim  of  =  A  <  a1',  V  >  x0. 

No- 
where now  V  is  any  number  >x0. 
Similarly, 

lim  a*  =  A'  >  a1,  J<Ot,, 

And  A<A'. 


420  CALCULUS 

Hence  as  before: 

0^,A'-A<al'-aK 

If  we  choose  I  and  V  both  as  rational  numbers  and  set  V  =  1+  h 
we  have : 

0^'  -A<al(ah-1), 

and  we  now  can  infer  as  in  the  earlier  proof  that  A'  =  A. 
It  remains,  therefore,  only  to  show  that  ax°  =  A.     Now  by  (1) 

ax  <  ax°  <  ax'        if        x<x0<  x\ 

Hence  lim  ax  ^  axo  ;g  lim  a*' 

*=x0-  x'=x0+ 

or  ^4  ^  a*o  g  J.. 

Thus  axo  is  seen  to  =  ^4,  and  this  completes  the  proof. 

We  have  hitherto  assumed  that  a  >  1.     It  is  shown  without 
difficulty  that  Theorems  1  and  2  hold  when  0  <  a  ?g  1. 

Theorem  3.     27ie  relations  (A)  q/*  (7/iap.  II,  §  8,  ^o/d  w/ien  m 
awe?  w  are  one  or  both  irrational. 

Consider,  for  example,  the  second  relation  : 

(am)n  =  amn. 

Let  m  approach  an  irrational  value,,  fi,  as  its  limit.  Then, 
since  xn  is  a  continuous  function  of  x  when  n  is  rational,  we 
have: 

lim  (aw)w=  (lim  am)tt=  (a*)n. 

»»==/*  m  =  /x  ' 

On  the  right-hand  side, 

limamn  =  a'An, 

m  =  jA 

and  hence 

(a»)n  =  a»n. 

If  here  we  allow  n  to  approach  an  irrational  number  v  as  its 
limit,  we  see  by  Theorems  1  and  2  that 


(«")" 


a^v. 


THE  EXPONENTIAL  FUNCTION  421 

The  proof  that 

(a™)"  =  amv 

depends  on  Theorem  1  alone. 

The  other  relations  of  (A)  are  proven  in  a  similar  manner. 

We  have  now  established  rigorously  all  that  was  assumed 
in  Chap.  IV  for  the  purpose  of  defining  the  logarithm  and  of 
differentiating  the  logarithm  and  the  exponential  functions. 
Hence  we  are  entitled  to  the  conclusion  of  that  chapter  that 
xn  is  continuous  and  has  a  derivative  when  n  is  irrational. 
We  have  also  the  material  for  proving  the  final  statements  of 
Chap.  II,  §  8,  respecting  the  graph  of  xn.  If  x  =  a,  (0  <  a  <  1 
or  a >  1)  and  y  =  b>0  are  chosen  arbitrarily,  one  and  only  one 
value  of  n  can  be  found  for  which  the  curve 

y  =  xn 

will  go  through  the  point  (a,  6),  namely  : 

b  =  a%  w  =  loga6  =  ^. 

log  a 

The  whole  subject  of  logarithms,  exponentials,  and  fractional 
exponents  can  be  treated  with  great  simplicity  by  basing  all 
of  these  functions  on  the  logarithm,  defined  as  the  definite 
integral : 

X 

dx 

x 


x 

/ 


Cf .  a  paper  by  Bradshaw,  Annals  of  Mathematics,  ser.  2,  vol.  4 
(1903),  p.  51 ;  or  Osgood,  Lehrbuch  der  Funktionentheorie,  vol.  1, 
p.  487. 


422  CALCULUS 


B.- FUNCTIONS   WITHOUT   DERIVATIVES 

In  recent  years  much  attention  has  been  paid  to  discontinu- 
ous functions  and  to  functions  which,  though  continuous,  still 
do  not  have  a  derivative.     Consider,  for  example,  the  function 

.    1 

y  =  sm  -  • 
x 

When  x  approaches  0  as  its  limit,  y  oscillates  between  the 
values  + 1  and  —  1,  and  thus  the  function,  while  remaining 
finite,  approaches  no  limit.  It  does  not  even  approach  one 
limit  when  x  approaches  0  from  the  positive  side  and  another 
limit  when  x  approaches  0  from  the  negative  side.  The  reader 
can  easily  plot  the  graph  roughly. 

Let  us  now  form  the  following  function : 

f(x)  =  x  sin-,        x^O; 
x 

/(0)  =  0. 

This  function  is  continuous  for  all  values  of  x,  and  its  graph 
is  comprised  between  the  lines  y  =  x  and  y  =  —  x.  At  the 
point  x  —  Oj  however,  it  has  no  derivative.  For,  form,  the 
difference-quotient : 

/(0+A*)-/(0)_dnl 
Ax  Ax 

This  variable  —  the  slope  of  a  secant  through  the  origin  and  a 

variable  point  P  with  the  coordinates  Ax  and  Ay  =  Ax  sin-- 

Ax 
oscillates  between  + 1  and  —  1,  i.e.  the  secant  OP  turns  to  and 
fro,  and  approaches  no  limit  whatever. 

Again,  a  function  may  have  a  first  derivative,  but  no  second 
derivative,  as  for  example: 

<f>(x)  =  x2sm-,         a?=£0; 
x 


*(<))  =  o. 


FUNCTIONS  WITHOUT   DERIVATIVES  423 


The  foregoing  functions  have  a  derivative,  to  be  sure,  in 
general ;  only  for  a  single  point  is  there  trouble.  But  exam- 
ples can  be  adduced  of  functions  that,  though  continuous  for 
all  values  of  x,  do  not  for  one  single  value  of  x  have  a  deriva- 
tive. 

In  the  light  of  these  facts  it  might  seem  as  if  a  thorough- 
going revision  of  all  we  have  said  in  the  early  chapters  were 
necessary.  The  revision,  however,  is  simple.  So  far  as  our 
theorems  about  derivatives  are  applied  to  special  functions  we 
have  fortified  ourselves  by  showing  that  the  elementary  func- 
tions actually  possess  derivatives  unless  possibly  at  exceptional 
points  easily  recognized.  In  the  statement  of  the  general 
theorems  of  Chap.  II,  §  4,  however,  it  is  true  that  we  need  to 
add  the  requirement  that  the  functions  u  and  v  shall  possess 
a  derivative.  With  this  supplementary  condition  Theorems 
I-V  are  true  in  all  cases.  The  proof  of  Theorem  V,  however, 
requires  a  modification,  of  which  we  will  speak  presently. 

Curves.  A  further  restriction  on  the  functions  we  have 
treated,  which  is  essential  for  some  of  the  proofs,  is  this,  that 
the  curve  y  —f{x)  shall  have  at  most  a  finite  number  of  max- 
ima and  minima  in  a  finite  interval.  The  function sf(x)  and 
<f>  (ic)  of  the  above  examples  do  not  have  this  property.  In  the 
neighborhood  of  the  point  x  =  0,  they  both  have  an  infinite 
number  of  maxima  and  minima.  We  can  impose  this  restric- 
tion, however,  throughout  the  Calculus  and  still  the  functions 
will  be  general  enough  for  most  purposes. 

With  this  restriction  the  proof  of  Theorem  V  is  valid. 
Without  it,  the  theorem  can  still  be  proven  by  the  aid  of  the 
Law  of  the  Mean. 

The  proof  of  convergence  required  to  justify  the  definition 
of  the  definite  integral,  Chap.  IX,  §  17,  rests  on  this  assumption. 


SUPPLEMENTARY  EXERCISES 
A.  —  Introduction 

Find  the  slope  of  each  of  the  following  curves. 

1.  y  =  x  —  x2,  x0  =  l.  Ans.   —  1. 

2.  y  =  4  -+-  x2,  x0  =  0.  Ans.   0. 

x0  =  i.  ^4ns.    —  9. 


Xq  —  Xq  . 

2aJ 


3. 

1-X 

y  = _, 

X 

4. 

2x 

y      (3  -  xfy 

5. 

x2  +  a2 

y  =  — 7—,  * 

6. 

a; 

"      3-2a 

8. 

a; 

0. 

a;     ar 

,    a?  =  sc0.   ^4ws.   1  — 


In-     6  + 

2x0 

^"*    (3- 

*o)3 

x2  +  2ax<i 

-a2 

Oo  +  a)2  (x0  +  a)- 

1 


7-  y=  1 

1  —  ar 

9.    2/  =  £3  —  2  £ 


a;-l 


r6—  1 

Differentiate  the  following  functions  by  the  Fundamental 
Method  of  p.  10. 

12.  2y  =  3x2-2x  +  l.  Am.   3x  -  1. 

13.  .y  =  a?4-3<B3  +  a!2-9a!-12.        Ans.   4ar5-9a;2  +  2a-9. 

14.  y  =  - 10  -12aj2  +  2aj8.  15.   y  —  x{2x  —  5). 

16.    y  =  x?(l  —  Sx).  17.    2/  =  5  — aa^  +  7aj. 

424 


SUPPLEMENTARY   EXERCISES  425 

2x 


»-\-x 

19. 

1 

20. 

y  =  xn. 

21. 

1 

V  =  — 

*      xm 

22. 

xy  +  x  =  ly. 

23. 

5x  +  2xy  —  4y  = 

=  3. 

24. 

ax+  b 

-dn«s. 


(1-z)2 

4 

-dns. , 

x5 

Arts.   nxn~x 


Ans.    -JjL 


.4WS. 


{i-xY 

7 


2  (2 -a)2 


ex  +  a  (cx  +  ay 

B.  —  Differentiation  of  Algebraic  Functions 

An  instructor  who  wishes,  after  finishing  Chapter  II,  to 
spend  more  time  on  pure  drill  work  in  differentiation  will 
probably  prefer  to  introduce  the  differential  at  this  point. 
This  may  be  done  by  taking  up  §§  1,  2,  4,  and  5  in  Chapter  V, 
omitting  in  §  5  all  that  relates  to  transcendental  functions.  A 
complete  set  of  the  formulas  needed  for  this  plan  (v.  Chap.  V, 
§  5)  is  herewith  appended. 

(A)         du  =  Dxudx         or         —  =  Dxu. 

dx 

I.  d(cu)  =  cdu. 

II.  d  (u  +  v)  =  du  +  dv. 

III.  d  (uv)  —  udv-{-v  du. 

TV  d  f^\  —  vdu  —  udv 


2. 


1.  dc  =  0. 

dxn  =  nxn~1dx        or 
dun  =  7iun~ldu. 


426  CALCULUS 

Differentiate  the  following  functions. 
25.   2o2-8a  +  3.  26.    Sx4-Sx2 -^-x-tt. 

27.    5«7-13^-9a;4-a;-f-l.      28.    a^-7a3a;-5a4. 
29.    (a +  10)  (5 -a)2.  30.  px2-(p  +  q)x-q. 

31.    0(2m  —  n$).  32.    (a  +  bx)(ax  —  b). 

33.  i£c3-2(a-6)»2  +  36a;-a  +  &. 

34.  |-2(3a-5)+z-l-^=^- 

35.  (3 -2a)7  36.    (p  +  g»)w. 

37.  a(2-3a)-4(l-a)2. 

38.  (2a  +  l)(l-2a)-3a(a  +  3)  +  l. 

39.  6 (5 -4a)3 -2(1 -2a)4.        40.    r4-4ar  +  a2. 

41.    3*2  -(a  -b)t  +  4a  -36.       42.    aw2  -  (2a+  (3)w- afi. 
43.   3a~4  — a^  +  a"1.  44.   4a5  —  \x~12  —  x~\ 

1_ 
7  a7 


45. 

2a-?- 

X 

47. 

1 

r • 

r 

49. 

2a  +  4 
a 

51. 

1  .  2 

a     a3 

53. 

7a2-5a  +  3 

a 

55. 

Z-f-  ma2 

57. 

58. 

x2_2cc  +  3 

46.    x2  +  ±-7. 


48. 
50. 
52. 
54. 
56. 


s 

x  —  x 

X 

'+V2.5 

1-a 

a2 

4 

7a2- 

5a  +  3 

8 

7ra4- 

-a  +  VlO 

x> 


Ans.    *=-£. 


59      *L=i 
-2a  59'    7-a3 


SUPPLEMENTARY  EXERCISES  427 

as2-2bs  +  c  aH      t2-l 


e 

62. 

1 

(l-*)2 

64. 

2x 

(5  -  xy 

66. 

2              1 

(4 -art*     4-a; 

67. 

a2-*2 
a  +  t 

69. 

14              3 

2-x     l+2a?      2-3a; 

71. 

12              4 

61. 

t 

63 

2 

(4 -3a;)3 

65. 

X2 

(a  +  bx)n 

Ans. 

68. 

2s 
1-s2' 

(4,-xf 


70.        a 


a-\-x     a  —  x 

72.     — £ — +  1. 

t  '  1-t2     2  +  3*2  1-s     s-s2^ 

g»(7  +  2aQ«  14^(24a?H-35)(7  +  2a;)3 

7       (14-  3a:)9  '  (14  -3a;)10 

75.       3* 


74. 

6x 

x2  —  x  -f- 1 

76. 

ar*-8 

tf  +  2x  +  ± 

77. 

a;4  —  a  + 1 

#2  +  0?  + 1 

79. 

a?  Va2  —  af . 

81. 

a; 

v2+l 

Ans.   1. 


78.  2a?       +      1       +  3s. 

(l-2a;)2l- 2a; 


80.    a;  -y/x2  +  a2. 


82      X?±^2- 
x  +  a 

a  +  x 


83.     Ja  +  a?-  84. 

\  a.  —  cf. 


a  —  x 


V2 


ax 


85.    (a2  +  a^)Va-a;.  86.    (*2  - 1)  VI  -  4*. 

3—  3/r-r^ 


87 


"&  88.    Jla± 

A/l  +  r  ^c  + 


c/^ 


428  CALCULUS 

89.     _A£±£_.  90.        «  +  ft* 


Va  —  /to  -V2ax  —  x2 

91.    arVA-ua;2.  92.    (/32  -  x2)Va  -  fix. 


93.    V4-5a;+arJ.  94.    V^6-5a;-a;2. 


95.    V-a;.  96.    aV4-5a?2. 

97.    rsV-l+3r-2r».  98.    *V^I 


99. 


t-t 


i_  .  100. 


Sx 


V^T?  (*  +  l)V4-3s» 


101.    a?  —  Vl+7a£  102.    Va2-^-Va2-t-£ 


103.  *  104.    VI-*-*. 

Va  -  a-—  Va  +  a;  Vl  —  a  +  a 

105. 


.        p  +  ^  +  1.  106_    J 


a;2_a._|_^  \  2  7-  cos  a 


107.  (2a-36a;)V(a  +  6a;)3.  ^4n«. —     -• 

108.  (8a2-12a^  +  15  6V)VFT^.^    iQSWV^+to 

: '         2 

109.  (2  a  —  6a;) Va  +  6a;. 

110.  (8  a2-  4  a&x  +  3  6W)  Va  +  6oj. 


111.    VC^-s2)3.  112.    V(a2  +  ^)3. 


113. 

V(m2  —  y2)5. 

115. 
117 

Vv2  —  1'02 
V 

4a;  +  17 

11Q 

Vl7-8ic-9a;2 

5— a;            ^5  +  10a; 

114.    V(p2+c2)5. 
116.    V^?. 
118.  ^^ 


V25  +  6a;-12a;2 
V-5  +  10aj-aj"^-10aj+^J'  600 


(_5  +  10a;-a;2)^ 


SUPPLEMENTARY   EXERCISES                     429 
120.  *  +  1         (n  +  iX+2^-  Ans.    2 


121. 

(Px-l)VP(l  +  x2)-2x 

2? 

n2        1                       1 

Ans.    V/o(l  +  x2) 

-2a; 

pi                  l 

*P     y/p(l+x2)-2x 

122. 

4a^  —  6x~^+x$* 

123. 

12aj*-20»*  +  8af*. 

124. 

-\/x 

125. 

1-* 

■Vt 

126. 

127. 

1  +  Vx 
y/x 

128. 

#?-Va?-f  2(^a>)«. 

An*.         2      +3Vi. 

3^a>        2 

129. 

(V») 

130. 

V2 

131. 

l  +  x+x2 
y/x 

132. 

x  —  x~l 

ViC 

133. 
135. 

(7  -9x)\/2x. 

1-r2 

Vmr 

134. 
136. 

Vax — » 

.2               |-3 

(a»  —  a;*)  . 

137. 

■y/x 

138. 
140. 

0-a 

y/sd 

139. 

V4-a2-5 
V2-a; 

141. 

m                  n 
naj  n  _  ma."»  _|_     w  _|_  ?l# 

142. 

cv1Al. 

143. 

(ZiC*  4-  ma6)2. 

144. 

»(l-af)* 

145. 

(a  +  x)(a2-x2)K 

146. 

(a  +  &af)F. 

147. 

(x*-xfi 

/4tt#,               2  or* 

(a?  •+.!)* 

(a? +  1)^-1)* 

430  CALCULUS 

148.  [V^=^-Vo+^]n.       149   [pii  +  x^-sxy. 

150. 


xn  —  x~n  151.  —  xn 


n 


m 

2 


152.    af+i  +  af-K  !53.    (xa  +  x  «)2. 


154. 


X1  1  KK       * 


155. 


i  +  yi-a2  x-Vi+x2 


157. 


Vi_vi  +  a/ 

%159. 

y(x-y)-x2  =  a. 

161. 

x  +  xy  =  a  +  b. 

163. 

6-V  +  «y  =  a262. 

165. 

aj2     y2  _  ^ 
a2     62 

.167. 

o     sc  —  y 

2/"= -• 

z  +  y 

XB6.    (_£_Y. 

158.  2«2  +  52/2=7. 

*  160.  xy~ax-\-by. 

162.  x3y  =  x03y0. 

164.  ar54-2/3-3aa^/  =  0. 

166.   y*  =  a2(x  —  y). 

168.    2^  —  3^  —  y2  =  l.  169.   a3  —  xy  +  y4  =  0. 

170.   aPy  —  y^  +  atf  —  aby^O.    171.   y2 —  2ax—2by -\-x*. 
172.   a4  +  2/4  +  l  —  6a&y  =  0.      173.    ^-3^  + 4a#2+lly7=l. 

C.  —  Differentiation  of  Transcendental  Functions 

The  further  special  formulas  of  p.  95  are  required,  which 
may  also  be  written  as  follows.  The  student  should  notice, 
however,  that  this  form  is  no  more  and  no  less  general  than 
that  of  p.  95. 

3.         dsinw      =  cos  udu. 


4. 

dcosu 

=  —  sin  udu. 

5. 

d  tan  u 

=  sec2  udu. 

6. 

dlogu 

__du 
u 

7. 

deu 

=  eudu. 

SUPPLEMENTARY  EXERCISES  431 

du 


8.         dsin-1w 


VI  -u2 


9.         a*tan_1w  = 


1  +  u2 


To  these  may  be  added,  if  desired : 

du 


10.         dcos~xu   = 


11.         c?vers_1w  = 


du 


■V2u-u2 
12.         daM  =  aMlogadw. 

The   student  should  note  further  the  trigonometrical  for- 
mulas :  * 

(A)  cos-1  u  =  -  —  sin-1  u. 

(B)  cot^u  =  tan~x  1  =  -  -  tan"1!*. 
w  u     2 

(C)  csc-1w  =  sin-1-. 

u 

(D)  sec-1  m  =  cos-1  -  • 

u 

Differentiate  the  following  functions.f 
174.    sin  ax.  175.    1  —  cos  irx.  176.    sin5 a;. 

177.    cos--  178.    csc2x.  179.    ^?- 

a  x 

180.    I  +  sme  181.    Sec(0-a).  182.    cotf. 

l-sm0  v         }  2 

183.    tan-^-.   '        184.    cos^^0"^-  185.    csc^- 
1  —  x  2  n 

*For  further  formulas   relating  to  the  trigonometric  functions,   cf. 
Peirce's  Tables,  pp.  73-80. 

f  Cf .  also  the  examples  on  pp.  96-99. 


432  CALCULUS 

186.        siD<^      ■  187.    1-°0S^.  188.    1  +  C0S^ 

1  —  cos  <j>  <f>  sin  <£ 

189.    cosn0.  190.    sinwirx.  191.    sinaxcos&x. 

192.    sin(n  —  m)6,      193.    cos  (n  +  ra)  <£.  194.    x2  cos  ax.  s 

.    (2n  +  l)x  .  f2n-l)x 

sin  V ! 1—  COS  x- 


sin  mx  2 


o 


195.    »"*•'"*.  196.    = 197.    - 

a?                                         ■    x  •    x 

•      X                                   sin-  sm- 

198.    vers  ax.  199.    xversx.              200.  vers2 ax. 


201.    Vl-A^sin^.    202.    xH  -»'        203. 


-45 

204.    sin  0  + vers  0.    2Q5     J    versa; 
sin  d  —  revs  0  *  2  —  rers 


COS  0  _™„  X 


cos  0  y  1  -J-  cos  X 


206. 


sin  0  —  vers  0  *  2  —  vers  x  Vvers  0 

.  207.    tang  - 1\        208.    cot  (Z  -  g.  %  209.    •  +  COtT*.  V 

.210.    cot  a;  — csc  a.       211.    cot--  .212.    cot  x  +  esc  x. 

£ 

«„„  o        cos3x  ,     •    o        A«j  r     i  sin7x  — cos9x 

213.    sm2x hsm3x.     214.    cos5xH 

3  63 

215.    sin2x  =  2sinxcosx.  216.    cos2x=  cos2x  — sin2x. 

»  217.  sin"1^^.       ■   218.  cos"1--  219.  tan-ll~2x. 

2  n  3 

220.  tan-1  (2  tan  x).  L  221.  cos-1  (n  sin  x).      222.  sin_1ax. 


223.  cot"1^— -•  224.    tan-^sinJ— 

a  +  x  V       ^2a 

ii) 


(  —  )•  226.    vers-1** 

a 

227.  sin-1^/-*  228.    vers"1  a 


6  a 


SUPPLEMENTARY   EXERCISES 


433 


229. 
232. 


tan" 


sin" 


,1l-2a; 

5a; 
x  b  +  a  cos  x 
a  +  b  cos  x 


230.  sin-1^^ 


231.  cos 


-il 


An$. 


233.  sin 


235. 

,236. 
239. 


_,  f  Va2  —  b2  sin  x~\ 
a  +  b  cos  a; 

^-.rva'-ysing-i 

6  +  a-  cos  a; 
log  (2  — 3a).        237.  log  (a— a), 
log  (1  +  x2).       «  240.  log  ^—^-  • 


234.    tan- 


-  Va2-&2 
a  -f  &  cos  a? 


'[>te>*"} 


^4ns. 


Va2-62 


.242.  log  VI  +  cos  6. 


♦  243.  log 


a  +  b  cos  x 
238.  logo2. 

241.  log  (a  +  2a2). 

Va  —  Va  —  x 


Va 


♦244. 
246. 
248. 
251. 
254. 
257. 

260. 
263. 

266. 
268. 
270. 

272. 
273. 


log  (a -a;)2. 


»+& 


.245.    e~x. 
^247.    el~x\ 


e*. 


log 


249.  exl°ea. 

252.  Va^. 

255.  emx  —  e~ 

258.  cP"*. 


e_nt  cos  (at  —  /3).  261.  e_xlogaa. 


ax  —  q~* 

az 
sin  log  a?. 

log(a  + VoM^)- 

^ein-1  x 


264.  alog<*». 


250.  a-x . 
253.   -Z/&. 
256.  log  log  a;. 
259.   (101+')2. 
262.  ez8ina:. 
265.  lO"*3. 


267.   logcosna. 
269.    ax  log  a. 
271.    VHK 


J_logVa  +  ^-Vat     a>a 

Va        Va  -f  6a  +  Va 

-**e*J*±Et    a<0. 
V-a  *    -« 

2f 


-4ws. 


^4ns. 


a;  Va  +  6a; 
1 


a;  Va  -f-  bx 


434  CALCULUS 


274.    log<*  +  Vtt2±*2 


275. 


togy 


V#  +  a  4-  V  #  —  a 


V#  +  a  —  V&  —  a 


276.    log(V21-8x  +  x2+  a -4). 


—  a 

scVa2  ±  or2 
1 

.4ns 

2Var>-a2 

1 

A  7is 

V21 

.  —  8  a;  -f  a,*2 
1 

277.  log(V-12  +  16a  +  4a>»  +  2a>  +  4).   Ans. 

"V  —  O  "T"4  X-\-X 
l 

278.  (sinaz)*.  279.   (l  +  x)\  280.   (log#)x\ 

~\tana;  i 

281.   (cos-1-  )      .        282.  (cos  a)*8.  283.  (4-tana;)x. 


a  J 

284.   (a2  +  xi)x.  285.   (tan-1a)8in_lx.       286.  rr2. 

i 
287.  (log  cos  x)x.  288.   (log  tan  x)x.         289.  xlogx. 

290.  x  =  Uel  +  e~~c).         Ans.   ^== — ,  or 


ec_e  c  V  a;       c 

291.  ex  cos^y  —(1  —  x)  log  (1  +  y)  =  1. 

Ans.  dy  =      **  CQS"ly  +  log  (1  +  y)      /I,     Wl^~ 
'  *»     e*Vl+2/+(l-a)Vr3^    +-^  ^ 

292.  r2  =  a2cos20.        293.  xy  —  yx  =  Q.         294.  0  =  sin(0  +  <£). 
295.  x  =  ysmy.  296.  2/  —  A:  sin  y  —  a  +  a;. 

297.  tan_1a?— tan"1?/ =  &#  +  &?/.    298.  #ey  —  ?/ log x  =  1. 
299.  asin(r#)  =  &0  +  r.  300.  rer  =  0sin0. 

301.  a£  +  ^  =  ai  302.  V^+V^=Va. 

3°3'  (a)   ~~(f)3  =  1'  304"   (1°g£C)2/-2/COS7raj  =  :I- 

{#  =  acos50,  (#  =  tan0  — 0, 

.    •    k»  306.    { 

y  =  osm^.  [y  =  cos6. 


SUPPLEMENTARY  EXERCISES  435 


(x  =  a  -f-  bt  +  ct* 
308. 
y  =  a'  +  b't  +  c't2. 


309.   i 


1-A2 

x  = -a. 

1+A2 

310. 

2a 
v  = a. 

*      1  +  A2 


1  +  *3 

x=-    -, 


2/  = 


311.  Find  the  slopes  of  the  curves  in  Exs.  308-310  where 
these  curves  cross  the  axis  of  x. 

312.  Remembering  that 

i r«+i 

l  +  r  +  r2+  ...  +rw=         r     , 
1  —  r 
show  that 

l+2r  +  3r2+  ...  +nr^  =  1  ~  (w  +  ^  +  <" 

(1-r)2 

313.  Using  the  result  of  the  preceding  question,  obtain  a 
formula  for  the  sum : 

l+22r  +  3V+  —  4-nV-1. 

314.  In  the  formula 

^     w-a      T(>  +  &)2' 

i2,  a,  6,  and  c  are  constants  and  p,  v,  and  T  denote  pressure, 
volume,  and  temperature,  respectively. 

(a)  Considering  T  as  constant,  find  the  derivative  of  p  with 
respect  to  v. 

(b)  Considering  v  as  constant,  find  the  derivative  of  p  with 
respect  to  T. 

(c)  Considering  p  as  constant,  find  the  derivative  of  T  with 
respect  to  v. 

D.  —  Applications 

Find  the  value  of  x  for  which  y  is  a  maximum  or  a  minimum 
in  each  of  the  following  cases. 

315.    12y  =  xs-6x\  x>0.  316.    ay  =  x7  -  35  x5,  x  <  0. 


436  CALCULUS 

317.   y  =  xB-  15a3 +  25,  x  >0.    318.    y  =  a?  -  2a?-  %x,  x>l. 

319.    2/  =  r-^— •  320.    y  = 


3  +  a;2  *      4  +  7(2-z)2 

321.    ,-*+!.  322.    r-£±* 

x  x 

323.    J.-13 -**  +  *,  ->3.      324.    »  =  |±*£,  *>3. 
cc  — 2  lx  —  6 

325.  Divide  12  into  two  such  parts  that  the  product  of  one 
of  these  parts  by  the  square  of  the  other  is  as  large  as  possible. 

326.  Divide  a  into  two  such  parts  that  the  product  of  one 
of  them  by  the  cube  of  the  other  is  as  large  as  possible. 

327.  Divide  30  into  two  such  parts  that  twice  the  cube  of 
one  of  them  increased  by  three  times  the  square  of  the  other 
may  be  as  large  as  possible. 

328.  If  the  strength  of  a  beam  is  proportional  to  its  breadth 
and  the  square  of  its  depth,  find  the  dimensions  of  the  strongest 
beam  that  can  be  cut  from  a  log  of  circular  cross-section. 

329.  What  is  the  shortest  distance  from  the  point  x  =  12, 
y  =  0  to  the  curve  y2  =  4:X? 

330.  What  point  of  the  curve  y  =  x%  is  nearest  to  the  point 
08  =  1,  y  =  0? 

331.  Tangents  are  drawn  to  the  parabola 

y2  =  a2  —  2  ax. 
What  one  of  them  cuts  off  the  smallest  triangle  from  the  first 
quadrant  ? 

332.  In  the  foregoing  problem,  what  tangent  has  the  short- 
est segment  in  the  first  quadrant  ? 

333.  A  trough  is  to  be  made  of  a  long  rectangular-shaped 
piece  of  copper  by  bending  up  the  edges  so  as  to  give  a  rect- 
angular cross-section.  How  deep  should  it  be  made,  in  order 
that  its  carrying  capacity  may  be  as  great  as  possible  ? 


SUPPLEMENTARY   EXERCISES  437 

334.  A  block  of  stone  is  to  be  drawn  up  an  inclined  plane 
by  a  rope.  Find  the  angle  that  the  rope  should  make  with  the 
plane  in  order  that  the  tension  may  be  as  small  as  possible. 

335.  Show  that  the  abscissa  of  that  point  of  the  curve 
y  =  log  x  which  is  nearest  to  the  origin  is  given  by  the  equa- 

tion:  iogi=*>. 

X 

336.  Find  the  value  of  x  from  the  foregoing  equation  correct 
to  two  significant  figures. 

Suggestion.  Tabulate  for  some  trial  values  of  x  the  values 
of  x2,  1/x,  log  1/x,  read  off  directly  from  convenient  tables, 
such  as  Huntington's  Four  Place  Tables. 

337.  Assuming  the  density  of  water  to  be  given  from  0°  to 
30°  C.  by  the  formula 

p=Po(l  +  at  +  l3t2  +  yt3), 

where  pQ  denotes  the  density  at  freezing,  t  the  temperature,  and 

a  =  5.30  x  10-5,         y8  =  -  6.53  x  10"6,         y  =  1.4  x  10"8, 

show  that  the  maximum  density  occurs  at  t  =  4.08°. 

Determine  by  inspection  whether  the  following  functions 
are  increasing  algebraically  or  decreasing  as  the  independent 
variable  increases. 

338.    a  —  x.  339.    t(t  —  c),  t>  c. 

340.    -i-,  x>-\.  341.    -1—,  o?>0. 

1  +  x  at  +  x2 

342.    x(x-l)(x-2),x>2.  343.    x(l  -  x2),  x>  1. 

344.    _L-,   *<0.  345.    l±i,  0<*<1. 

2  2 

346.    a  ~X  .  0<x<a.  347.    -^— ,  0<x<a. 

as  +  x*  a  —  x 

348.    Xs  —  bx,  6<0.  349.    t(l+t2). 

350.    (x^2)(x-7)(x-20),  x<2.    351.   a4  +  5  a2*2  +  7  .-c4. 


•      2  +  Sx 

354. 

5  —  x 

u= 

4  + 2a; 

356. 

2x              A 

1  -\-ar 

358. 

y  =  x  —  a?  +  x4,  x  —  1. 

360. 

y  =  x  sin  x,  x  =  tt. 

438  CALCULUS 

Are  the  following  functions  increasing  or  decreasing  ? 

-     «-4  +  *-  353.    .-I±l!. 

4  +  5* 

355.    y^  +  fo- 

y  +  Sz 

357.   2/  =  ^^,  »-fc 

359.    s  =  100*-16*2,  Z  =  3. 
361.    y  =  x  —  2  sin  a?,  a;  =  0. 
362.   y  =  2  sinx  —  3  cos  2x,  x=  \ir.   363.    y  =  a; log x,  x  =  1. 

In  what  intervals  are  the  following  functions  increasing,  and 
in  what  decreasing  ? 

364.   y  =  10-3x-{-xz.  365.  u  =  —  5+30a?  —  20a?. 

366.    y  =  x4  —  10a;3  —  41.  367.    «  =  t8  +  «*  +  8. 

368.   y  =  2<e8-15a2  +  36a>-l.       369.   r  =  3P  +  50-ll. 

370.    y  =    9  X      .  371.   y  =  a;  +  sina;. 

a2  +  a,*2 

372.  y  =  sin  a;  -f-  cos  x.  373.   y  =  e~xsin  x. 

374.  a?  =  A  sin  (n«  —  y),  A  >  0,  n  >  0. 

375.  aj=(7cos(w«  +  €),     C  >  0,  n  >  0. 

376.  x  =  e_a'  sin  (n£  —  e),  n  >  0. 

In  what  intervals  are  the  following  curves  concave  upward, 
and  in  what  downward  ? 

377.  y  =  x?-9x2  +  7x-3.  378.    y  =  2ar? +  3^-1. 
379.    y  =  3x5  +  5x4-2x  +  7.           380.   y  =  x11  - xM. 
381.   y  =  ax3+bx2  +  cx  +  d.             382.   y  =  ax2+bx  +  c. 

383.    y  =  2z6-9ari  +  10a:4-a;.        384.    y  =  — -^  +  — • 
B  y     56      42^30 

385.    y  =  x-|-cosa;.  386.    y  =  sina?  — a;. 


SUPPLEMENTARY  EXERCISES  439 

387.   y  =  x\ogx.  388.    y  =  xe~x. 

389.    y  =  ™%£.  390.    y  =  xx. 

x 

391.    y  =  e~x\  392.    y  =  logcosx. 

393.  Plot  the  curves  which  represent  the  functions  in  Exs. 
384-392. 

394.  A  point  moves  along  the  hyperbola  xy  =  100  and  its 
ordinate  increases  uniformly  at  the  rate  of  20  ft.  a  sec.  Find 
how  fast  the  abscissa  is  decreasing  when  y  =  15. 

395.  A  point  moves  along  the  curve  r=l/0  at  the  rate  of 
6  ft.  a  sec.     How  fast  is  radius  vector  turning  when  0  =  2tt  ? 

396.  If  a  drop  of  rain,  as  it  moves  through  moist  air,  receives 
accretions  so  that  its  radius  is  increasing  at  the  rate  of  c  cm.  a 
sec,  at  what  rate  is  its  volume  increasing  when  its  radius  is 
a  cm.? 

397.  A  point  describes  the  cardioid  r  =  2a(l  —  cos#)  with 
uniform  velocity  c.  How  fast  is  its  distance  from  the  pole 
r  =  0  changing  when  9  =  \-k  ? 

398.  A  man  in  a  train  that  is  running  at  full  speed  looks 
out  of  the  window  in  a  direction  perpendicular  to  the  track. 
If  he  fixes  his  attention  successively  for  short  intervals  of 
time  on  objects  at  different  distances  from  the  train,  show  that 
the  rate  at  which  he  has  to  turn  his  eyes  to  follow  a  given 
object  is  inversely  proportional  to  its  distance  from  him. 

399.  A  point  describes  the  circle  x2  -f-  y2  =  25  with  a  velocity 
of  12  ft.  a  sec.  Find  the  component  velocity  along  the  axis  of 
x  when  x  =  4. 

400.  At  what  rate  is  the  ordinate  of  the  curve  y  —  x—x^ 
changing  when  x=  1,  if  the  abscissa  is  increasing  at  the  rate  of 
10  metres  a  second  ?    Is  the  ordinate  increasing  or  decreasing? 

401.  A  man  walks  across  the  floor  of  a  semicircular  rotunda 
100  ft.  in  diameter,  his  speed  being  4  ft.  a  sec,  and  his  path 
the  radius  perpendicular  to  the  diameter  joining  the  extremi- 


440  .       CALCULUS 

ties  of  the  semicircle.  There  is  a  light  at  one  of  the  latter 
points.  Find  how  fast  the  man's  shadow  is  moving  along  the 
wall  of  the  rotunda  when  he  is  half  way  across. 

402.  Water  is  flowing  out  of  a  hemispherical  bowl  from  an 
opening  at  the  lowest  point.  If  the  rate  of  efflux  is  c  cu.  cm.  a 
sec.  when  the  level  of  the  water  is  half  way  between  the  hole 
and  the  centre  of  the  bowl,  how  fast  is  the  level  falling  ? 

Suggestion.  Compare  the  volume  of  water  that  flows  out 
in  At  seconds  with  the  fall  in  level,  Ax,  and  thus  compute 
directly  lim  Ax/ At. 

E. —  Errors  of  Observation* 

Let  x  be  an  observed  quantity  and  y  a  second  quantity, 
dependent  on  x,  to  be  computed: 

y  =/(*)• 

An  error  of  Ax  in  observing  x,  the  true  value  being  x0,  will 
give  rise  to  an  error  Ay  in  the  computed  value,  where 

ty  =/(^o  +  A^)  —f(x0). 

Now  we  are  concerned  only  with  an  approximate  value  of  Ay 
and  hence  any  other  quantity  that  differs  from  Ay  by  less  than 
the  error  in  Ay  which  we  are  willing  to  admit  is  an  equally 
faithful  representative  of  the  error  in  y.  Such  a  representative 
is  found  for  most  of  the  cases  that  arise  in  practice  in  dy, 
since,  if  the  derivative  f'(x)  is  finite  for  x  =  x0  and  =£0,  Ay 
and  dy  will  differ  from  each  other  only  by  a  small  percentage 
of  either,  when  Ax  is  small,  that  is : 

limAy-<fr  =  0. 

A*=o      dy 

Definition.  —  By  the  absolute  error  in  y  is  meant 
dy=f'(x0)dx, 

*I  am  indebted  to  Dr.  Harvey  N.  Davis  for  the  problems  in  this 
section. 


SUPPLEMENTARY   EXERCISES  441 

the  error  in  x  being  denoted  by  dx.     The  relative  error  is 
defined  as  dy 

J' 

Since  the  relative  error  is  d  log  y,  it  is  often  better  to  take 
the  logarithm  of  each  side  of  the  equation  y=f(x)  before 
differentiating. 

If  y  depends  on  several  variables,  x,  t,  •••,  it  is  frequently- 
desirable  to  consider  the  errors  in  y  arising  separately  from 
the  errors  in  x,  t,  •••;  i.e.  to  hold  all  but  one  of  the  letters 
x,  t,  "-  fast  and  allow  that  one  alone  to  vary. 

By  the  coefficient  of  propagation  is  meant  the  multiplier  A. 
when  the  error  equation  is  written  in  one  of  the  four  standard 

du  dx  dx        dii 

dy  =  \xdx.      —  =  A.2 — ,       dy  —  k3 — ,       —  =  \4dx. 
*  y  x  *  x  y        M 

Thus,   when   both   errors  are  absolute  errors,   A.=A.1=/'(a?0); 

when  both  errors  are  relative  errors,  A.  =  A2  =  a,0/f(#0)//(a,'0); 

and  so  on. 

Example.     The  length  and  the  diameter  of  a  cylindrical  bar 

are  nearly  25  cm.  and  2  cm.,  respectively.     Find  the  absolute 

and  the  relative  errors  in  the  volume  due  to  an  error  of  8  =  .02 

cm.  in  measuring  the  diameter. 

Here,  V=±ir&H. 

Hence  the  absolute  error  is 

dV=  i-rrDHS  =  1.6  cu.  cm. 

The  relative  error  may  be  found  by  dividing  each  side  of 
the  equation  by  F=^7rZ)2^Z"; 

—  -2--.02, 

or  2  per  cent.    Had  the  relative  error  been  the  only  thing  to  be 
computed,  we  should  have  proceeded  more  simply  as  follows : 

logF=logi,r#+21ogZ), 

dV  =  2dD  =  28 
V         D       D' 


442  CALCULUS 

403.  In  the  case  of  a  sphere  find  the  absolute  error  in  the 
volume  produced  by  an  error  of  .01  cm.  in  the  radius,  r, 
(a)  when  r  =  1  cm.  j    (6)  when  r  =  50  cm. 

Ans.    (a)  .126  cu.  cm. ;    (b)  314  cu.  cm. 

404.  In  the  two  cases  of  the  preceding  problem,  find  the 
relative  error  in  the  volume  due  to  an  error  of  2  per  cent  in 
determining  r.  Ans.    6  per  cent  in  both  cases. 

405.  What  is  the  allowable  absolute  error  (a)  in  the  meas- 
urement of  the  longest,  and  (b)  in  the  measurement  of  the 
shortest  dimension  of  a  rectangular  block  10  cm.  by  5  cm.  by 
2  cm.,  if  its  volume  is  to  be  determined  within  one-fifth  of  one 
per  cent  ?  Assume  in  each  case  that  the  other  measurements 
are  correct.  Ans.    (a)  .02  cm. ;    (6)  .004  cm. 

406.  What  is  the  allowable  relative  error  in  measuring 
(a)  the  diameter  of  the  base  (5  cm.)  and  (b)  the  height  (10  cm.) 
of  a  right  circular  cone,  if  its  volume  is  to  be  determined  within 
a  fifth  of  one  per  cent  ?  Assume  in  each  case  that  the  other 
measurement  is  correct. 

407.  What  conditions  must  the  dimensions  of  a  right  cylin- 
der satisfy  if,  for  a  given  error  in  the  volume,  the  allowable 
absolute  errors  in  the  length  are  equal  ? 

408.  A  desired  quantity,  S,  is  the  sum  (or  difference)  of  two 
measured  quantities,  a  and  b.  Show  that  the  coefficients  of 
propagation  for  relative  errors  are  numerically  as  the  quantities 
a  and  6,  and  that  the  allowable  relative  errors  are  inversely  as 
a  and  b. 

409.  What  is  the  relation  between  the  allowable  absolute 
errors  in  the  last  question  ? 

410.  If  y=a  —  b  and  3a  =  4&, .what  relative  error  in  y  is 
caused  by  an  error  of  one  per  cent  in  a  ? 

411.  A  flag  pole  subtends,  at  a  point  40  ft.  from  its  base,  an 
angle  of  60°.  What  relative  error  in  the  computed  height  is 
caused  by  an  error  of  1°  in  the  observed  angle  ? 


SUPPLEMENTARY  EXERCISES  443 

412.  What  relative  error  is  caused  by  an  error  of  one  per 
cent  in  the  distance  ? 

413.  Would  the  computed  height  in  Ex.  411  be  more  or  less 
sensitive  to  the  error  in  the  observed  angle  if  the  observer 
moved  farther  away  from  the  pole  ? 

414.  At  what  distance  from  the  pole  is  the  computed  height 
least  sensitive  to  errors  in  the  observed  angle  ? 

415.  If,  in  Ex.  411,  a  minute  of  arc  in  the  angle  and  a  tenth 
of  one  per  cent  in  the  distance  from  the  foot  of  the  pole  are 
degrees  of  accuracy  in  the  measurement  about  equally  easy  to 
obtain,  where  should  the  observer  stand  ? 

416.  Discuss  the  accuracy  of  the  determination  of  the  ordi- 
nate of  a  point  on  a  given  circle  centred  at  the  origin  when  the 
abscissa  is  measured. 

417.  A  surveyor  has  a  measured  base  line  of  100  miles  and 
finds  that  the  angles  between  it  and  a  distant  mark  are  60°  and 
45°.  With  what  absolute  accuracy  must  each  of  the  three 
measurements  be  made  if  no  one  of  the  resulting  errors  in  the 
shorter  of  the  unknown  distances  is  to  exceed  one  hundredth  of 
one  per  cent  ? 

418.  A  steel  cylinder  is  8  cm.  long  and  6  cm.  in  diameter, 
and  it  weighs  20  gr.  The  moment  of  inertia  of  such  a  cylinder 
about  its  geometric  axis  being 

what  is  the  allowable  absolute  error  in  the  measured  diameter 
if  /  is  desired  within  one  hundredth  of  one  per  cent  ? 

419.  The  moment  of  inertia  of  a  right  cylinder  about  a  line 
through  its  centre  perpendicular  to  the  axis  of  figure  is 

where  I  is  the  whole  length.     If  the  diameter  of  the  cylinder 


444  CALCULUS 

in  Ex.  418  is  known  to  one  part  in  600,  with  what  relative  ac- 
curacy should  I  and  M  be  known,  to  give  equally  small  errors 
in  J? 

420.  A  certain  magnetic  measurement  (in  Gauss's  "  B  posi- 
tion ")  leads  to  the  formula 

and  in  a  certain  case  s  was  35  cm.,  I  was  4  cm.,  and  a  was  4°. 
If  M/H  is  to  be  determined  within  a  fifth  of  one  per  cent, 
what  is  the  allowable  absolute  error  (a)  in  s,  and  (b)  in  I  ? 

421.  A  similar  magnetic  measurement  (in  Gauss's  "  A  posi- 
tion ")  leads  to  the  formula 

^(^i^tana, 
H  2s 

and  in  a  certain  case  s  was  35  cm.,  I  was  4  cm.,  and  a  was  8°. 
What  relative  error  in  M/H  would  be  caused  by  an  error  of 
three  hundredths  of  one  per  cent  in  s?  What  is  the  cor- 
responding relative  error  in  I  ? 

422.  When  a  magnet  or  pendulum  is  swinging  through  a 
viscous  medium,  like  air,  it  is  found  that  the  successive  ampli- 
tudes form  a  geometric  series :  that  is,  if  any  amplitude  be 
called  a0,  those  following  are 

a1  =  a0/k,     a2=a0/k2,     •     •     •  ,     am  =  a0/lcm, 
where  k  >  1  is  a  constant  of  the  apparatus.     If  log  k  (called  the 
"  logarithmic  decrement ")  be  denoted  by  A.,  then 
A  =  log  op  -  log  am 

s  m 

Assuming  that  a0  is  correct,  but  that  every  subsequent  ampli- 
tude am  is  subject  to  an  absolute  error  8  independent  of  m,  find 
the  absolute  error  in  X  in  terms  of  a0,  k,  and  m,  and  show  that 
it  is  a  minimum  when  m  satisfies  the  equation  km  —  e  =  2.718, 
so  that  under  the  conditions  of  this  problem  one  should  use, 
in  determining  X,  that  am  which  is  nearest  a0/2.718  to  get  the 
best  results. 


SUPPLEMENTARY   EXERCISES  445 

423.  Assuming  that  in  the  preceding  problem  both  a0  and 
am  are  subject  to  the  absolute  error  8,  and  that  the  resulting 
error  in  A  is  the  sum  of  the  absolute  values  of  the  errors  which 
would  be  produced  by  the  errors  in  a0  and  am  acting  separately, 
show  that  the  best  value  of  m  is  given  by  the  equation 

424.  Solve  the  foregoing  equation  by  approximate  methods 
and  show  that  km  or  a0/am  should  have  the  value  3.59. 

425.  In  Ex.  423  assume  that  the  error  in  A  has  its  "most 
probable  value,"  which  is  the  square  root  of  the  sum  of  the 
squares  of  the  errors  which  would  be  produced  by  the  errors 
in  a0  and  in  am  acting  separately.  Show  that  the  best  value 
for  aQ/am  is  now  3.03. 

426.  The  period  of  a  pendulum  varies  inversely  as  the 
square  root  of  the  force  of  gravity,  g,  at  the  place  of  observa- 
tion. With  what  percentage  accuracy  must  the  period  be 
observed  if  g  is  to  be  determined  to  one  part  in  ten  thousand  ? 

427. *  If  the  period  of  such  a  pendulum  is  very  nearly  one 
second,  the  period  can  be  compared  with  that  of  the  pendulum 
of  a  clock  beating  seconds.  Suppose  that  the  two  are  beating 
exactly  together  at  the  time  tlf  that  the  unknown  pendulum 
gains  on  the  other,  and  that  they  beat  together  again  at  the 
time  t2.  Then  in  exactly  n  =  t2  —  tx  sec.  the  unknown  pendulum 
has  made  exactly  n  -\- 1  swings,  and  its  period  is  n/(n  + 1)  sec. 
How  large  must  n  be  to  make  an  error  of  1  sec.  in  determining 
n  allowable  under  the  conditions  of  the  preceding  question  ? 

428.  If  the  n  of  the  last  question  is  45,  what  percentage 
error  in  the  period  would  result  from  an  error  of  2  sec.  in 
determining  n? 

429.  If  the  allowable  error  in  the  period  is  a  tenth  of  one 
per  cent,  how  great  must  n  be  to  make  an  error  in  it  of  10  sec. 
allowable  under  the  conditions  of  Ex.  427  ? 

*  The  solutions  of  Exs.  427-429  belong  in  a  course  in  physics,  for  the 
mathematical  processes  involved  are  exceedingly  simple.  They  are  inserted 
merely  to  indicate  more  clearly  the  bearing  of  such  work  as  is  here  given 
in  errors  of  observation,  on  the  actual  problems  that  arise  in  practice. 


446  CALCULUS 

F.  —  Integration 
Integrate  each  of  the  following  functions  with  respect  to  x. 
430.   12x5-10xi-16x?  +  2x.     431.    a/-5a>6  +  a?  +  l. 

M 

432.    axh  —  bxa  +  ab.  433.    (m  +  n)xn  +  m  —  n. 

434.    1- 4a -or5 +  H #10.  435.     -  3  +  x  +  £«*-  \x*. 

436.    4a?9-13x6-5ar3-l.  437.    1^-1^  +  2^  +  1. 

438.    9-3"  +  "2.  439.    *-g*  +  *-l. 

12  3x 

440.    ic2+--l.  441.    3--  +  03. 

a?  a; 

442.    a£  +  af£  +  l.  443.    a^  +  af*  — 2. 

444.    VaJ  — -— +  -•  445.    af*  —  a"1  +  x~K 

■y/x      x 

446.    -* ?-- i  +  a  447.     £L±^±£^. 

448.    9** -4 -4"  449.     iri£. 

450.    1  +  ^g-(l  +  a?)Vg.  451.    x(a+&V2a?). 

a? 

452.    (5  — VaJ)(a>  +  3Va>).  453.    (a  +  bx)^/cx-. 

454.    (Vc-V2x)2.  455.    [-{/a5  — -g-. 

\  wax. 

456.    (a2  -  x2)  VTax.  457.    (Va-V3to)3. 

Evaluate  the  following  integrals. 

458-  J(r^-ir->-     459-  fiihi-^*- 

460.  /•-*_.  461.  rj^p i=L 

J  Vl-ay  J   \     Va  L  +  £ 

*        a. 


SUPPLEMENTARY  EXERCISES  447 


462.      f**L.  463.      ffjLL ?«_U 

J  a-x  J  \a  +  t     a-3tj 

464.       /*±-?fe  465.       fr-W-«-lfc 
7466.     f  (aam2x-bcos3x)dx.   467.     je-a2xdx. 

4e8-  /(^I-1)^-  469-  f(j=i—J* 

470.      rfetC0Sa-~\dt.  471.      i*—^—dx. 

Jtf2.      Ccot-dx.  S     ^473.      Csm  —  dx. 

474.      I  sec  ax  dx.  475.     fcsc^dx. 

476.     fsec22xdx.  477.      /7l  +  cot2  |^d «. 

478.      j  cos  (nt-e)dt.  479.      /V*~ 

/'        o             o  i  /iai          (*CSC2xdx 

tansxsec2xdx.  M  45i*     I — , * 
J    Vcota? 


"tan" 

a2-}-^2 


r  -i     da  48^/  / 

482.      I  cos  Jaj —            «  / 

J        vr^2"  ^ 

484.v7excose*cta.  J  485.     J  e~x(l- e~x)3dx. 

486.1JVF+2**  48V^  J^pT 

488.      Pt{$F* -■*■"*)&.  V^89'      f^a*Jr^dx. 


448  CALCULUS 

490.     f(a  -  3 bx)%dx.  491.  /  V"^  dx,  {x  <  0). 

492.      C   ^ads    .  493.  f(k-r)m~ndr. 

J  Vc  +  3as  J 

494.      C(Z±^dr.  495.  C(a2 -2ax)^dx. 

[(axy^-iaxy^dx.     497.  j  ^7=' 

r^tan  —  da?.  499.  AinS^Sdft 

J  3         3  J  Va 


498. 


500.      I  a  cos 


502 


504 


ra  cos  _2g-3a  ote.  501 .     T-Va  sin  ( Va  *)  dt. 

/kao       /"2 cos aa;— sin  flsc, 
(2 e-  -  c^«) dr.  503-   J  g  +  /8  daj- 

.      C(a  +  b-axfdx.  ■     505.      C(a  +  t)*to. 

•  Jv(ife.-4)^        507-  /(rf-«- 


4    cto. 


V  508.     fx(a2-x>)*dx.  509.   j?V~a  +  b-afidt. 

510.  /•_**=.  511.  A^*   • 

512.     fx(l+aa?)idx.  513.      ix(a  +  bx2) 

514.      A**  515.      /"     * 

Jl  +  ct6  J3«- 


dic. 


516 


46a-5 


SUPPLEMENTARY   EXERCISES  449 

518.      Ilosx  — '  V619.      /  x  tan  irx2 dx. 


520 


522 


524 


526. 


•    Jlogx^'  Uld.    J  a 

.     jxe-^dx.  Vfel.     flog^/1—xdx. 

.     jx\ogVa2-x2dx.  523.      Cxf—^-^-b^dx. 

.    C  **  +  **>.  525.    r  (b-^dx    . 

J  x2  +px  +  q  J  (a  +  3bx-  Xs)2 

fap\pl  _{]-%.  527.      f       2atdt 

r_l^_.  529.      f-3* 

J5-2*2  J5  +  2 


540 


542 


2  a;2' 


530.       f J* -•  531.       f_*- 

J  (x  +  a)*  +  (x-a)*  Jl  +  7i 

532.      f-*L.  533.      f- 

J  ?-i>2  J  63  +  27  s2 

534.      /*- -^ -.  535.      f 

J  (x  +  a)2-(x-af  J  i 


2ds 

+  2 

dq 


m2  -j-  n2q2 

dx 

'x 


536.       A™*^.  537.      A1^ 

,/     sin2  a;  ^z     cos5: 

538.      f-^-v  539.       f * ■ 

J  p  +  qu2  J  b  +  (x  —  a)2 

f        cfa  541       f    (34-5^)0 


5V* 


.      f -***-•  543.      f-^JL. 
J  1  +  x*  J  m-y* 

544.    f   X'dx    .  545.    fj^*. 

J5  +  13*16  J  1-x2 


2g 


450  CALCULUS 


546.      f*xldx  547<     jtB,n(nx-2)dx. 

J  Vlog*  J 

548.      /V-^dr.  549.    JV~ -(n  +  2*), 

cos  a?  VI  —  &2  sin2 a? dec.    551.      I  -« 

^  cos2- 
i 

/C    sin  OdO 
e-»in9 cos  Odd.  553.      /      ,-  =' 

J  V4  — costf 


e?as 
as 


554.  f 'sin  (9  log  cos  OdO.  555.  /cos  log  a;  ^ 

556.  n  +  cosj^  557#  fC0S2esindde. 

J   1-COS0  J 

558.  fsmxco$L2xdx.  559.  /  cos  *  sin  2  *  da*. 

560.  /sin2*  cos  a- d*.  561.  I  sin2 a;  cos2 *  d*. 

562.  /  sinw*cos*d*.  563.  I  cosn*sin*d*. 

564.  I  cos2 mxdx.  565.  I  sm2mxdx. 


V 


566. 


/* 


d*  ^  1  *     _i/&tan* 


^4?is.    — ■  tan' 


cos2*-f-&2sin2*  ct& 


567       f    sinftcosftda;    .    ^s.    1 log  (a  cos2* +  6  sin2*). 

J  a  cos2* +  5  sin2*  2(&-a) 

C      cos*d* 
568.       /  

J  Vl  —  ft2  sin*  a; 

/,         sin  (m  —  w)  *      sin  (m  +  fl)  * 

smm*smn*d*.  <4n*.       2\m_J!) 2(m  +  w) 


SUPPLEMENTARY  EXERCISES 

„_-.       C  .                     1        a  cos  (m  —  n)x     cos  (m-f  n)x 

570.  I  smmxcosnxax.     Ans. — ■? f — ) — ! — f-- 

J  2(m  —  ri)            2  (m  -f  n) 

.»..        C                       j  a        sin(m  —  n)x  ,  sin(m-fw)a; 

571.  I  cos  mx  cos  na;  dx.  ins.    ——) f-  -\ — ) ' — f- . 

J  2  (ra  —  ti)           2(m  +  %) 

Evaluate  the  following  ;  grals  by  the  aid  of  the  Tables. 

572.      C ^ .  573.      T— — — 

J3~7x  +  2xi  j3-2z  +  7(* 

/*                    _7_       /*        asdar 

"    J  3s             f-Saj8"  '   j3-2x  +  7a*' 

576        C——^—~  577        C——^—— 

'    Jl3-7x  +  2x2y  J  3aj9-2ar,  +  7q!* 

578.  r    ^  579.  r — *» — 

J  V3-7a>  +  2a*  J  V3-2a;  +  7arJ 

580.       f-            '         =-  581.      /^ * 

J  xV8-7o;  +  2it*2  J  #V3-2o;  +  7arJ 

582.       f ** 583.       f *» -• 

J  (3_7^  +  2.t2)^  J  (3- 2a; +  7^* 

f— ^ 585.     f- 

J  4  —  5  cos  a;  ^7  5 


584. 


586. 


588. 


590. 


592. 


5--4cosa; 
dx 


r   Mx  r 

J  3  +  Ttanx'  '   J  11  + 

f * 589.      f 

J  (5  —  4  cos  a;)2  ^7  < 

r * 59i.  r 

J  sin  x  (5  —  4  cos  a?)  j  1 

f fe 593.       f-J^ 

J  10  —  cos  a;  +  2  sin  a;  J1  +  cc 


13  sin  x 
dx 


cos  #(5  —  4cos#) 

dx 

9  —  7  cos2ic 


dx 

COSiC 


452  CALCULUS 

G.  —  Definite  Integrals 

594.  The  hyperbola  xy  —  lOOl  rotates  about  the  axis  of  x. 
Find  the  volume  of  that  part  of  the  solid  thus  generated  which 
is  contained  between  the  planes  ^rpendicular  to  the  axis  and 
corresponding  to  x  =  5  and  x  =  2 

595.  The  curve  y  =  sec  x  revolvt  '"♦out  the  axis  of  x.  Find 
the  volume  of  the  solid  whose  bases  ;  ,rrespond  to  a?  =  \v  and 

596.  The  curve  y  =  x  —  xA  rotates  about  the  axis  of  x.  Find 
the  volume  of  the  solid  generated  bytthat  part  of  it  which  lies 
above  the  axis  of  x. 


597.    The  hyperbola       ^_?f 

«2    i>2 


=  1 


revolves  about  the  axis  of  x.  Find  the  volume  cut  off  from 
one  of  the  two#  solids  thus  obtained  by  a  plane  perpendicular 
to  the  axis  and  distant  h  from  the  vertex. 

Ans>?¥¥(3a  +  h). 

3d2  J 

598.  So  much  of  that  arc  of  the  curve  y  =  eosx  —  icos2# 
which  cuts  the  axis  of  ordinates  and  lies  above  the  axis  of  x 
rotates  about  the  latter  axis.  Find  the  volume  of  the  solid 
generated. 

599.  The  curve  y  =  cos_1ic  rotates  about  the  axis  of  x.  Find 
the  volume  generated  by  that  part  of  the  curve  for  which 
0  <^  y  <  ir,  the  base  being  a  plane  perpendicular  to  the  a: 

|  =  -1. 

600.  The  parabola  x1  +  y1  =  a'£  rotates  about  the  axis  of  x. 
Find  the  volume  of  the  solid  bounded  by  the  arc  which  is  tan- 
gent to  the  coordinate  axes  at  its  extremities,  the  base  being 
formed  by  a  plane  through  the  origin  perpendicular  to  the  axis. 

601.  Determine  the  volume  of  the  following  solid.  Think 
of  the  axis  of  y  as  vertical  and  consider  the  cylinder  whose 


SUPPLEMENTARY  EXERCISES  453 

elements  are  perpendicular  to  the  curve  y  =  12  — 12  a£  Next, 
turn  this  cylinder  about  the  axis  of  y  through  90°.  The  two 
cylinders  and  a  horizontal  plane  through  the  origin  bound  the 
solid  in  question. 

602.  If  the  base  of  the  conoid  of  Ex.  5,  p.  161,  is  an  ellipse 
whose  plane  is  parallel  to  the  fixed  line,  show  that  the  volume 
is  ^irabh,  where  h  denotes  the  distance  from  the  line  to  the 
plane. 

603.  The  solid  of  p.  159,  Fig.  49,  is  cut  by  a  plane  through 
O,  perpendicular  to  the  plane  of  the  base  AOB  and  making  an 
angle  of  45°  with  OA.  Determine  the  volume  of  the  part 
with  the  vertex  A. 

604.  A  horn  is  generated  by  a  variable  circle  whose  plane 
turns  about  a  fixed  line.  The  point  of  the  circle  nearest  the 
line  describes  a  quadrant  AB  of  a  circle  of  radius  a,  and  the 
radius  of  the  variable  circle  is  cO,  where  0  denotes  the  angle 
between  the  variable  plane  and  its  initial  position,  when  it 
passes  through  A.  Show  that  the  volume  of  the  horn  is 
j^v^iSa  +  Sirc). 

605.  Find  the  area  of  the  lateral  surface  of  the  solid 
described  in  Ex.  603. 

606.  Find  the  area  of  the  lateral  surface  of  the  solid 
described  in  Ex.  601.  Ans.    64,  nearly. 

607.  An  arbitrary  closed  curve  is  drawn  on  the  surface  of  a 
sphere,  catting  out  a  region  S  from  that  surface.  Show  that 
the  volume  of  the  cone  whose  vertex  is  at  the  centre  of  the 
sphere  and  whose  base  is  S  is 

where  A  denotes  the  area  of  S  and  R  the  radius  of  the  sphere. 

608.  The  curve  r=f($)  rotates  about  the  axis  0  =  0.  As- 
suming f(6)  to  be  single-valued  and  continuous  for  a  _  0  _  /3, 
where  0  _  a  <  /3^v,  obtain  a  formula  for  the  volume  of  the 
solid  generated  by  the  rotation  of  the  plane  region  bounded 
by  the  curve  and  the  two  radii  vectores  drawn  to  its  extremities. 


454  CALCULUS 

609.  Hence  determine  the  volume  of  the  solid  generated  by 
the  rotation 

(a)  of  the  curve  r  =  a  cos  20; 

(b)  of  the  lemniscate  ?*2  =  a2  cos  2  6  j 

(c)  of  the  curve  r  =  1  —  62. 

610.  Show  that,  if  two  solids  are  so  related  to  each  other 
that,  when  cut  by  any  plane  parallel  to  a  certain  fixed  plane, 
the  areas  of  the  two  cross-sections  are  equal,  then  the  volumes 
of  the  solids  are  equal.     (Cavalieri's  Theorem.) 

611.  Find  the  areas  of  the  surfaces  in  (a)  Ex.  594;  (b)  Ex. 
597. 

612.  Find  the  fluid  pressure  on  the  vertical  plane  area 
bounded  by  the  curve  a-, 

a4  +  ar 

and  the  double  ordinate  x  =  h,  the  axis  of  y  lying  in  the  sur- 
face of  the  liquid. 

613.  Assuming  that  the  density  of  water  at  a  distance  of 
x  ft.  below  the  surface  is 

p  =  p0(l  +  .  000  001 3  x), 

find  how  much  greater  the  pressure  is  on  a  vertical  rectangle 
10  ft.  broad  and  a  mile  deep,  with  one  side  in  the  surface,  than 
what  it  would  be  if  water  were  incompressible. 

614.  Find  the  pressure  on  the  end  of  the  trough  described 
in  Ex.  2,  p.  164,  if  the  density  is  a  linear  function  of  the  dis- 
tance below  the  surface. 

615.  If  the  density  p  of  any  curve  is  variable,  show  that  the 
mass  of  the  curve  is  t 

M—  I  pds. 


/< 


616.    The  density  of  a  rod  is  proportional  to  the  distance 
from  one  end.     Find  its  mass. 


SUPPLEMENTARY  EXERCISES  455 

617.  The  density  of  a  semicircular  wire  is  proportional  to 
the  perpendicular  distance  from  the  diameter  joining  its  ends. 
Find  its  mass. 

618.  If  in  the  preceding  problem  the  density  is  proportional 
to  the  square  of  the  perpendicular  distance  from  the  radius 
drawn  perpendicular  to  the  above  diameter,  what  is  the  mass  ? 

619.  Find  the  centre  of  gravity  of  the  rod  in  Ex.  616. 

620.  Find  the  centre  of  gravity  of  a  quadrant  of  the  wire  in 
Ex.  617  and  in  Ex.  618. 

621.  The  density  of  a  spherical  surface  at  any  point  is  pro- 
portional to  the  distance  of  the  point  from  a  fixed  diameter. 
Eequired  the  mass. 

622.  The  same  problem  for  a  cone  of  revolution,  the  density 
being  proportional  to  the  distance  from  the  axis. 

623.  The  density  at  each  point  of  a  sphere  is  proportional  to 
the  distance  of  the  point  from  the  centre.     Find  its  mass. 

624.  If  the  density  is  a  -f  6r,  where  r  denotes  the  distance 
from  the  centre,  required  the  mass. 

Determine  the  following  moments  of  inertia  and  radii  of 
gyration : 

625.  A  rod  whose  density  is  proportional  to  the  distance 
from  one  end,  about  a  perpendicular  at  that  end. 

626.  The  same  rod,  about  a  perpendicular  bisector. 

627.  The  circular  wire  of  Ex.  617  about  the  diameter. 

628.  The  same,  about  the  radius  perpendicular  to  the 
diameter. 

629.  The  circular  wire  of  Ex.  618  about  the  diameter. 

630.  The  same,  about  the  radius  perpendicular  to  the 
diameter. 

631.  A  circular  disk  whose  density  is  proportional  to  the 
distance  from  the  centre,  about  the  centre. 


456  CALCULUS 


632.    The  same,  about  a  diameter. 


633.  A  circular  disk  whose  density  is  proportional  to  the  dis- 
tance from  the  circumference,  about  the  centre. 

634.  The  same,  about  a  diameter. 


635.  A  circular  disk  whose  density  at  any  point  is  \vb2  —  r2, 
r  denoting  the  distance  from  the  centre,  about  the  centre. 

636.  The  same,  about  a  diameter. 

637.  A  conical  surface  of  revolution,  about  the  axis  of  the 
cone. 

638.  A  spherical  surface,  about  a  diameter. 

639.  A  conical  surface  of  revolution  whose  density  is  pro- 
portional to  the  distance  from  the  axis,  about  the  axis. 

640.  A  spherical  surface  whose  density  is  proportional  to 
the  distance  from  a  diameter,  about  that  diameter. 

641.  A  sphere  whose  density  is  proportional  to  the  distance 
from  the  centre,  about  a  diameter. 

642.  A  sphere  whose  density  is  proportional  to  the  distance 
from  a  diameter,  about  that  diameter. 

643.  A  sphere  whose  density  is  any  linear  function  of  the 
distance  from  the  centre,  about  a  diameter. 

644.  A  triangle  whose  density  is  proportional  to  the  dis- 
tance from  one  side,  about  that  side. 

645.  A  semicircle  whose  density  is  proportional  to  the  dis- 
tance from  the  bounding  diameter,  about  that  diameter. 

Determine  the  following  centres  of  gravity : 

646.  A  uniform  circular  segment. 

647.  A  uniform  circular  sector.     Check  your  answer. 

648.  A  segment  of  the  equilateral  hyperbola 

x2  —  y2  —  a2, 
cut  off  by  the  double  ordinate  x  =  a  +  h. 


SUPPLEMENTARY   EXERCISES  457 

649.  The  corresponding  segment  for  any  hyperbola. 

650.  A  triangle  whose  density  is  proportional  to  the  dis- 
tance from  one  side. 

651.  A  uniform  parabolic  wire,  extending  equal  distances  to 
each  side  of  the  vertex. 

652.  A  semicircular  wire  whose  density  is  proportional  to 
the  length  of  the  arc  measured  from  its  middle  point. 

653.  The  same,  when  the  density  is  proportional  to  the  dis- 
tance from  the  diameter  through  its  extremities. 

654.  A  semicircle  whose  density  is  proportional  to  the  dis- 
tance from  the  bounding  diameter. 

655.  A  semicircle  whose  density  is  proportional  to  the  dis- 
tance from  the  centre.  (Suggestion.  First  obtain  a  formula 
for  the  centre  of  gravity  of  a  semicircle  whose  density  is  an 
arbitrary  function  of  the  distance  from  the  centre.) 

656.  Show  that  the  ordinate  of  the  centre  of  gravity  of  the 
uniform  plane  area  of  §  1,  p.  153,  is  given  by  the  formula :  * 

6 

I  y2dx 

657.  Find  the  attraction  of  a  quadrant  of  a  circle  on  a 
particle  at  the  centre  of  the  circle. 

658.  Find  the  attraction  of  so  much  of  a  cylindrical  surface 
of  revolution  as  lies  between  two  planes  normal  to  the  axis,  on 
a  particle  situated  in  the  axis. 

659.  The  same,  when  the  density  of  the  surface  is  pro- 
portional to  the  distance  from  one  of  the  planes. 

660.  Find  the  attraction  of  a  homogeneous  hemispherical  sur- 
face on  a  particle  situated  at  the  centre  of  the  sphere. 

*  This  formula  was  given  to  me  by  Mr.  Rogers  Sherman  Hoar,  at  that 
time  a  student  in  the  first  course  in  the  Calculus. 


458  CALCULUS 

661.  The  same  question,  only  that  the  density  of  the  surface 
is  proportional  to  the  distance  from  the  axis. 

662.  The  same,  when  the  density  is  proportional  to  the 
distance  from  the  base  measured  along  the  arc  of  a  great  circle 
meeting  the  base  at  right  angles. 

663.  Find  the  attraction  of  the  semicircular  wire  (a)  in  Ex. 
617 ;  (6)  in  Ex.  618,  on  a  particle  at  the  centre  of  the  circle. 

664.  The  same  for  a  particle  on  the  circumference  extended, 
at  the  point  situated  symmetrically  with  respect  to  the  wire. 

665.  Find  the  attraction  of  a  homogeneous  surface  in  the 
form  of  a  right  cone  of  revolution  on  a  particle  at  the  centre 
of  the  base. 

666.  Find  the  attraction  of  the  surface  of  a  frustum  of  a 
cone  of  revolution  on  a  particle  at  the  centre  of  the  smaller 
base. 

667.  Evaluate  the  double  integral 


// 


xydS, 


where  S  is  the  rectangle  whose  vertices  lie  at  the  points  (1,  2), 
(1,  5),  (3,  2),  (3,  5).  Ans.  42. 

668.  The  same  integral,  extended  over  the  triangle  cut  off 
from  the  first  quadrant  by  the  line  joining  the  points  (0,  3) 
and  (3,  0).  Ans.  3f . 

669.  Compute 


C  C(±o+x+f)ds, 


the  region  S  being  the  piece  of  the  plane  bounded  by  the 
parabola  y  =  x2  —  x  and  the  right  line  y  —  x.  Ans.   55^. 


670.    Extend  the  integral 

//WHO" 


SUPPLEMENTARY   EXERCISES  459 

over  the  same  region,  and  check  your  answer  by  inverting  the 
order  of  integration. 

671.   Compute  the  value  of 


// 


<x?ydS, 


the  region  S  consisting  of  a  triangle  whose  vertices  are  at  the 
origin  and  the  points  (1,  2)  and  (2,  1). 

672.  Check  your  result  in  the  last  question  by  using  polar 
coordinates  skilfully. 

673.  Find  the  centre  of  gravity  of  the  solid  consisting  of 
the  part  cut  out  of  a  homogeneous  sphere  by  two  planes 
through  the  centre. 

674.  Show  that  the  moment  of  inertia  of  a  homogeneous 
right  cylinder  about  an  axis  through  its  centre  perpendicular 
to  its  axis  of  figure  is  /n£      n  \ 

/=Jf(i+l2> 

675.  Find  the  moment  of  inertia  of  a  homogeneous  right 
cone  of  revolution  about  an  axis  through  the  vertex  perpen- 
dicular to  the  axis  of  figure. 


INDEX 


Absolute  value,  5. 
Acceleration,  190. 
Addition  theorem,  75. 
Arasler's  plani meter,  409. 
Approximate      computations,     Chap. 

XX. 
Area  under  a  curve,  111,  117,  153,  407 ; 

—  in  polar  coordinates,  132  ; 

—  of  a  surface  of  revolution,  167 ; 

—  of  any  surface,  367 ; 

—  of  cylindrical  surface,  370; 
further  problems  in  areas,  453. 

Attraction  of  gravitation,  181,  457. 

Cardioid,  152. 

Catenary,  131,400. 

Caustics,  350. 

Centre  of  fluid  pressure,  175. 

Centre  of  gravity,  169 ; 

general  formulation  for  — ,  174, 381 ; 

further  problems  in  — ,  456. 
Change  of  variable,  295. 
Chart,  A  Mercator's,  331. 
Compound  interest  law,  82. 
Computation,  Numerical,  Chap.  XX. 
Concave  upward,  Test  for,  50; 

further  examples  for  — ,  438. 
Confocal  quadrics,  326. 
Conservative  field  of  force,  395. 
Contact  of  curves,  276. 
Continuity,  4,  20,  283. 
Convergence,  Fundamental    principle 

for,  246. 
Cubic  equation,  58,  61. 
Curvature,  Definition  of,  134; 

radius  of  — ,  136 ; 

formula  for  —  in  polar  coordinates, 
145. 
Curve  tracing,  51 ; 

further  problems  in  — ,  110,  439. 


Curves  on  the  sphere,   cylinder,   and 
cone,  329; 

—  without  tangents,  422. 
Cycloid,  146. 

Definite  integral,  156,  358,  380; 

further  problems  relating  to  — ,  452. 
Density,  Examples  of  variable,  183, 

188,  360,  376,  381,  396,  454. 
Derivative,  9; 

logarithmic  — ,  22 ; 

partial  — ,  287 ; 

directional  — ,  308; 

normal  — ,  309. 
Developments  in  series,  Chap.  XIII. 
Differentials,  Definition  of,  91 ; 

—  of  higher  order,  94; 

—  of  arc,  99 ; 

—  of  functions  of  several  variables, 
293; 

total  — ,  293 ; 
exact  — ,  309. 
Differentiation,  General  formulas  of, 
13,21,94; 
special  formulas  of  — ,  95 ; 

—  of  implicit  functions,  34,  302. 
Direction  cosines,  283,  289,  319. 
Directional  derivatives,  308. 
Double  integral,  358,  458. 
Duhamel's  theorem,  164. 

Elementary  functions,  96. 

Envelopes,  344. 

Epicycloid,  149. 

Equiangular  spiral,  91,  130. 

Errors  of  observation,  etc.,  306,  440. 

Euler's     theorem    for    homogeneous 

functions,  300. 
Evolnte,  139,  349. 
Exact  differentials,  309. 


460 


INDEX 


461 


Fluid  pressure,  161. 
Four-cusped  hypocycloid,  150. 
Fractional  exponents,  27. 
Function,  Definition  of  a,  2; 

continuous  — ,  4,  20,  58 ; 

implicit  — ,  34,  302 ; 

algebraic  —,35; 

test  for    increasing    or    decreasing 
-49; 

further  examples  of  latter,  281,  437 ; 

inverse  — ,  68 ; 

elementary  —,96; 

—  of  several  variables,  282 ; 
hyperbolic  — ,  412 ; 
exponential  — ,  417 ; 

—  without  a  derivative,  422. 

Gudermannian,  416. 
Guldin,  c/.  Pappus. 

Helix,  318. 

Homogeneous  functions,  300. 
Hyperbolic  functions,  412. 
Hypocycloid,  150. 

Implicit  functions,  34,  302. 
Indefinite  integral,  156. 
Indeterminate  forms,  231,  278,  280. 
Indicator  diagrams,  Area  of,  409.  , 

Infinitesimals,  85; 

—  with  several  variables,  293. 
Infinity,  19. 

Inflection,  Point  of,  51,  53,  276. 
Integral,    Definition    of    (indefinite), 
114; 
tables  of  — ',  126; 

—  as  the  limit  of  a  sum,  155, 358,  380 ; 
definite  — ,  156,  358,  380 ; 

iterated  — ,  354,  382 ; 

double  — ,  358 ; 

surface  — ,  374 ; 

triple  — ,  380 ; 

line  — ,  390 ; 

numerical   computation  of  definite 

—,406; 
further    problems    in    double    and 

triple  — ,  458. 
Integration,  Special  formulas  of,  118; 
further  problems  in  — ,  446 ; 
variable  limits  of  — ,  186 ; 
mechanical  — ,  409. 
Iterated  integral,  354,  382. 


Jacobian  determinant,  305. 

Law  of  the  mean,  230,  334; 

generalized  — ,  234. 
Laws  of  motion,  190. 
Lemniscate,  133. 
Length  of  a  curve,  129,  166. 
Limits,  Three  theorems  about,  15; 

the  —  %,  etc.,  231,  278,  280; 

—  for  functions  of  several  variables, 
282. 

Line  integrals,  390. 

Logarithmic  spiral,  cf.   Equiangular 

spiral. 
Loxodrome,  330. 

Mass  in  terms  of  density,  360,  381. 
Maxima  and  minima,  39,  53,  275,  336; 

further  problems  in  — ,  104,  279,  435. 
Mercator's  chart,  331. 
Minima,  cf.  Maxima. 
Moment  of  inertia,  176,  359,  381 ; 

further  problems  in  — ,  455. 

Napierian  logarithm,  Reason  for,  82. 
Natural  logarithm,  cf.  Napierian  — . 
Newton's  ba/ws  of  motion,  1'K); 

—  method  for  numerical  equations, 
399. 

Normal,  Equation  of,  37; 

—  lines  and  planes,  286, 289, 316, 317 ; 

—  derivative,  309; 
principal  — ,  325. 

Numerical  computation,  Chap.  XX. 

Osculating  circle,  138,  277  ; 

—  plane,  323. 

Pappus,  Theorems  of,  362. 
Peirce's  Tables,  126. 
Planimeter,  Amsler's,  409. 
Principal  normal,  325. 

Quadratic  forms,  340. 

Quadric  surfaces,  Confocal,  326. 


Radian  measure,  Reason  for, 
Radius  of  gyration,  178. 
Rates,  cf.  Velocity. 
Rhumb  "line,  330. 
Rolle's  theorem,  229. 


66. 


462 


INDEX 


Roots  of  equations,  57,  398; 

further  problems  in  — ,  110, 281 ,  322. 
Roulettes,  150. 

Semi-cubical  parabola,  141. 

Series,   Definition  of  infinite,  244; 

power  -  - ,  257 ; 

Maclaurin's  —  ,  202 ; 

Taylor's  —  ,  264,  266,  335 ; 

binomial  —  ,  272. 
Simple  harmonic  motion,  201. 
Simpson's  rule,  406. 
Slope  of  a  curve,  5. 
Spirals,  131 ; 

—  of  Archimedes,  145 ; 

also  Equiangular  spiral. 
Square  root  sign,  4. 
Successive  approximations,  403. 
Surface  integral,  374. 

Tables  of  integrals,  126. 
Tangent,  Slope  «tf,  5,  11 ; 


Tangent,  Equation  of,  37 ; 

—  in  polar  coordinates,  90 ; 

—  plane,  289,  316 ; 

—  line  to  space  curve,  317. 
Taylor's  series,  264 ; 

—  theorem,  266,  3:35. 
Test-ratio,  250. 
Total  differential,  292. 
Triple  integral,  .380,  459. 
Trisectrix,  145. 
Trochoids,  151. 

Velocity,  Definition  of,  46 ; 

rates  and  —  ,  101 ; 

further  problems  in  — ,  108,  109,  439. 
Volume  of  a  solid  of  revolution,  157; 

volumes  of  other  solids,   159,   352; 

further  problems  relating  to  such 
volumes,  452. 

Work  done  by  a  variable  force,  393, 
409. 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

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